Area Under Curves

Calculus, 2ADV EQ-Bank 22

The diagram shows the graph of \(y=\log _e(x+1)\)

Express \(x\) as a function of \(y\). (1 mark)

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Hence, or otherwise, find the exact area of the shaded region bounded by the curve \(y=\log _e(x+1)\), the \(x\)-axis, and the line \(x=3\). (3 marks)

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a. \(x=e^y-1\)

b. \(A=4 \ln 4-3 \ \text{u}^2\)

Show Worked Solution

a. \(y=\ln (x+1) \ \Rightarrow \ x+1=e^y \ \Rightarrow \ x=e^y-1\)

b. \(\text{Area of rectangle}=\ln 4 \times 3=3 \ln 4\)

\(\text{Find the area between curve and \(y\)-axis from \(\ y=0\ \) to \(\ y=\ln 4\):}\)

\(A\)	\(=\displaystyle \int_0^{\ln 4} e^y-1\, d y\)
	\(=\Big[e^y-y\Big]_0^{\ln 4}\)
	\(=\left(e^{\ln 4}-\ln 4\right)-(1)\)
	\(=4-\ln 4-1\)
	\(=3-\ln 4\)

\(\text{Shaded Area}\)	\(=3 \ln 4-(3-\ln 4)\)
	\(=4 \ln 4-3 \ \text{u}^2\)

Calculus, 2ADV EQ-Bank 29

The diagram shows the graph of \(y=\log _2 2 x\)

Determine the exact value of the shaded area bounded by the \(x\)-axis, the \(y\)-axis, and the curve \(y=\log _2 2 x\).

Express your answer is the form \(\dfrac{a}{\ln b}\), where \(a\) and \(b\) are integers. (4 marks)

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\(A=\dfrac{7}{\ln 4}\ \text{u}^2\)

Show Worked Solution

\(y=\log _2(2 x) \ \Rightarrow \ 2 x=2^y \ \Rightarrow \ x=\dfrac{1}{2} \times 2^y\)

\(A\)	\(=\dfrac{1}{2} \displaystyle \int_0^3 2^y\, d y\)
	\(=\dfrac{1}{2}\left[\dfrac{2^y}{\ln 2}\right]_0^3\)
	\(=\dfrac{1}{2}\left[\dfrac{2^3}{\ln 2}-\dfrac{1}{\ln 2}\right]\)
	\(=\dfrac{7}{2 \ln 2}\)
	\(=\dfrac{7}{\ln 4}\ \text{u}^2\)

Calculus, 2ADV C4 EQ-Bank 19

The diagram shows the graph of \(y=\ln (x+2)\).

Find the exact value of the shaded area bounded by the \(y\)-axis, the line \(y=\ln 6\) and the curve \(y=\ln (x+2)\). Express your answer in the form \(a+b\,\ln c\) where \(a, b\) and \(c\) are integers. (4 marks)

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\(A=(4-2 \ln 3) \ \text{u}^2\)

Show Worked Solution

\(y=\ln (x+2)\ \ \Rightarrow\ \ x+2=e^y\ \ \Rightarrow\ \ x=e^y-2\)

\(y\text{-intercept occurs at}\ (0,\ln 2).\)

\(A\)	\(=\displaystyle \int_{\ln 2}^{\ln 6}\left(e^y-2\right) d y\)
	\(=\Big[e^y-2 y\Big]_{\ln 2}^{\ln 6}\)
	\(=e^{\ln 6}-2 \ln 6-e^{\ln 2}+2 \ln 2\)
	\(=6-2-2\left(\ln \dfrac{6}{2}\right)\)
	\(=(4-2 \ln 3) \ \text{u}^2\)

Calculus, 2ADV C4 2025 MET2 17 MC

Given that \(\displaystyle \int_1^2 f(x)\, d x>\int_1^3 f(x)\, d x\), the graph of \(y=f(x)\) could be

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\(A\)

Show Worked Solution

\(\displaystyle \int_1^2 f(x)\, d x>\int_1^3 f(x)\, d x\)

\(\text{To be true,} \ \displaystyle \int_2^3 f(x)\, d x<0 \ \ \text{(“net” area below} \ x \text{-axis)}\)

\(\Rightarrow A\)

♦ Mean mark 38%.

Calculus, 2ADV C4 2025 HSC 25

Show that \(\dfrac{d}{d x}(\sin x-x\, \cos x)=x\, \sin x\). (2 marks)

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Hence, find the value of \(\displaystyle\int_0^{2025 \pi} x\, \sin x \, dx\). (2 marks)

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The regions bounded by the \(x\)-axis and the graph of \(y=x\, \sin x\) for \(x \geq 0\) are shown.

Let \(A_n=\displaystyle \int_{(n-1) \pi}^{n \pi} x\, \sin x \,dx\), where \(n\) is a positive integer.
It can be shown that \(\left|A_n\right|=(2 n-1) \pi\). (Do NOT prove this.)
Find the exact total area of the regions bounded by the curve \(y=x \sin x\), and the \(x\)-axis between \(x=0\) and \(x=2025 \pi\). (2 marks)
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a. \(\text{See Worked Solutions}\)

b. \(2025 \pi\)

c. \(4\,100\,625 \pi \ \text{units}^2\)

Show Worked Solution

a.	\(\dfrac{d}{dx}(\sin x-x\, \cos x)\)	\(=\dfrac{d}{dx} \sin x-\dfrac{d}{dx} x\, \cos x\)
		\(=\cos x+x\, \sin x-\cos x\)
		\(=x\, \sin x\)

b.	\(\displaystyle\int_0^{2025 \pi} x\, \sin x\)	\(=\Big[\sin x-x\, \cos x\Big]_0^{2025 \pi}\)
		\(=\Big[(\sin (2025\pi)-2025 \pi \times \cos (2025 \pi))-0\Big]\)
		\(=0-2025 \pi \times-1\)
		\(=2025 \pi\)

c. \(\text{Area}=\displaystyle \int_0^\pi x\, \sin x \, dx+\left|\int_\pi^{2 \pi} x\, \sin x \, dx\right|+\cdots+\int_{2024 \pi}^{2025 \pi} x\, \sin x \, dx\)

\(\text{Using}\ \ \left|A_n\right|=(2n-1) \pi:\)

\(A_1=\pi, A_2=3 \pi, A_3=5 \pi, \ldots, A_{2025}=4049 \pi\)

\begin{aligned}
\rule{0pt}{2.5ex} \text{Area}& =\underbrace{\pi+3 \pi+5 \pi+\ldots+4049 \pi}_{\text {AP where } a=\pi, \ l=4049 \pi, \ n=2025} \\
\rule{0pt}{4.5ex} & =\frac{2025}{2}(\pi+4049 \pi) \\
\rule{0pt}{3.5ex} & =4\,100\,625 \pi \ \text{units }^2
\end{aligned}

♦♦ Mean mark (c) 32%.

Calculus, 2ADV C4 EQ-Bank 26

The graph of \(f(x)\) is drawn below

Evaluate \(\displaystyle \int_0^6 f(x)\, d x\) (2 marks)

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Evaluate \(\displaystyle \int_0^6[f(x)-3]\, d x\) (2 marks)

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Evaluate \(\displaystyle \int_4^6 f^{\prime}(x)\, d x\) (1 mark)

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a. \(8 \dfrac{1}{2}\)

b. \(-9 \dfrac{1}{2}\)

c. \(-3\)

Show Worked Solution

\(\displaystyle \int_0^6 f(x)\)	\(=\text{net area above the}\ x \text{-axis }\)
	\(=\text{Area 1}+ \text{Area 3}-\text{Area 2}\)
	\(=\left(6+1 \dfrac{1}{2}\right)+3-2\)
	\(=8 \dfrac{1}{2}\)

b. \(f(x)-3 \ \ \text{shifts graph (above) 3 units lower:}\)

\(\displaystyle\int_0^6[f(x)-3] \ \ \text{will be negative (areas below \(x\)-axis)}\)
	\(=-\left(1 \dfrac{1}{2}+5+3\right)\)
	\(=-9 \dfrac{1}{2}\)

c.	\(\displaystyle \int_4^6 f^{\prime}(x) d x\)	\(=[f(x)]_4^6\)
		\(=f(6)-f(4)\)
		\(=0-3\)
		\(=-3\)

Calculus, 2ADV C4 EQ-Bank 30

Evaluate \(\displaystyle \int_{-2}^0 \sqrt{4-x^2} \, d x\). (3 marks)

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\(\pi\)

Show Worked Solution

\(\sqrt{4-x^2} \ \ \text{is a semicircle, centre}\ \ (0,0) \ \ \text {and radius 2.}\)

STRATEGY: This integral is beyond 2ADV integration techniques. An alternate strategy is required.

\(\displaystyle \int_{-2}^0 \sqrt{4-x^2} \, d x \ \)	\( \ =\ \text{Shaded area (above)}\)
	\(=\dfrac{1}{4} \times \pi+2^2\)
	\(=\pi\)

Calculus, 2ADV C4 EQ-Bank 31

Evaluate \(\displaystyle \int_0^5\abs{x^2-4 x+3} dx\). (3 marks)

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\(\dfrac{28}{3}\)

Show Worked Solution

\(\text{Shaded Area}\)

\(=\displaystyle \int_0^5\abs{x^2-4 x+3} dx\)

\(=\left[\dfrac{x^3}{3}-2 x^2+3 x\right]_0^1+\left|\left[\dfrac{x^3}{3}-2 x^2+3 x\right]_1^3\right|+\left[\dfrac{x^3}{3}-2 x^2+3 x\right]_3^5\)

\(=\left(\dfrac{1}{3}-2+3\right)+\left|(9-18+9)-\left(\dfrac{1}{3}-2+3\right)\right|+ …\)

\(\left[\left(\dfrac{125}{3}-50+15\right)-(9-18+9)\right]\)

\(=\dfrac{4}{3}+\left|-\dfrac{4}{3}\right|+\dfrac{20}{3}\)

\(=\dfrac{28}{3}\)

Calculus, 2ADV C4 EQ-Bank 6 MC

The value of \(\displaystyle \int_0^6\abs{x-2}\,dx\) is

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\(B\)

Show Worked Solution

\(\displaystyle \int_0^6\abs{x-2}\,dx=\dfrac{1}{2}(2 \times 2)+\dfrac{1}{2}(4 \times 4)=10\)

\(\Rightarrow B\)

Calculus, 2ADV C4 2024 MET2 4*

If \( { \displaystyle \int_a^b f(x) d x=-5 } \) and \( { \displaystyle \int_a^c f(x) d x=3 } \), where \(a<b<c\).

Find \( { \displaystyle \int_b^c 2 f(x) d x } \). (2 marks)

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\(16\)

Show Worked Solution

\({ \displaystyle \int_a^b f(x) d x} +{ \displaystyle \int_b^c f(x) d x}\)	\(={ \displaystyle \int_a^c f(x) d x} \)
\(-5+{ \displaystyle \int_b^c f(x) d x}\)	\(=3\)
\({ \displaystyle \int_b^c f(x) d x}\)	\(=3+5=8\)

\(\therefore{ \displaystyle \int_b^c 2f(x) d x}=16\)

Calculus, 2ADV C4 2024 HSC 14

The curves `y=(x-1)^2` and `y=5-x^2` intersect at two points, as shown in the diagram.

Find the `x`-coordinates of the points of intersection of the two curves. (1 mark)

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Find the area enclosed by the two curves. (3 marks)

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a. `x=2\ \text{and}\ -1`

b. `9\ \text{units}^2`

Show Worked Solution

a. `y=(x-1)^2, \ y=5-x^2`

\(\text{Intersection occurs when:}\)

`(x-1)^2`	`=5-x^2`
`x^2-2x+1`	`=5-x^2`
`2x^2-2x-4`	`=0`
`2(x-2)(x+1)`	`=0`

`x=2\ \text{and}\ -1`

b.	`\text{Area}`	`= \int_{-1}^{2} (5-x^2)-(x-1)^2\ dx`
		`=\int_{-1}^{2} 5-x^2-x^2+2x-1\ dx`
		`=\int_{-1}^{2}4-2x^2+2x\ dx`
		`=[4x-\frac{2}{3}x^3+x^2]_{-1}^{2}`
		`=[(8-\frac{16}{3}+4)-(-4+\frac{2}{3}+1)]`
		`=\frac{20}{3}-(-\frac{7}{3})`
		`=9\ \text{u}^2`

Calculus, 2ADV C4 2023 MET2 6 MC

Suppose that \(\displaystyle \int_{3}^{10} f(x)\,dx=C\) and \(\displaystyle \int_{7}^{10} f(x)\,dx=D\). The value of \(\displaystyle \int_{7}^{3} f(x)\,dx\) is

\(C+D\)
\(C+D-3\)
\(C-D\)
\(D-C\)

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\(D\)

Show Worked Solution

\(\text{Given }\displaystyle \int_{3}^{10} f(x)\,dx=C\ \ \text{and}\ \displaystyle \int_{7}^{10} f(x)\,dx=D\)

\(\text{We can deduce:}\)

\(\displaystyle \int_{3}^{10} f(x)\,dx\)	\(=\displaystyle \int_{3}^{7} f(x)\,dx+\displaystyle \int_{7}^{10} f(x)\,dx\)
\(C\)	\(=\displaystyle \int_{3}^{7} f(x)\,dx+D\)
\(C-D\)	\(=\displaystyle \int_{3}^{7} f(x)\,dx\)
\(\therefore\ \displaystyle \int_{7}^{3} f(x)\,dx\)	\(=D-C\)

\(\Rightarrow D\)

♦ Mean mark 49%.
MARKER’S COMMENT: 41% of students chose option C incorrectly assuming \(\int_{7}^3 f(x)\,dx=\int_{3}^7 f(x)\,dx\).

Calculus, 2ADV C4 2023 HSC 32

The curves \(y=e^{-2 x}\) and \(y=e^{-x}-\dfrac{1}{4}\) intersect at exactly one point as shown in the diagram. The point of intersection has coordinates \(\left(\ln 2, \dfrac{1}{4}\right)\). (Do NOT prove this.)

Show that the area bounded by the two curves and the \(y\)-axis, as shaded in the diagram, is \(\dfrac{1}{4} \ln 2-\dfrac{1}{8}\). (3 marks)

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Find the values of \(k\) such that the curves \(y=e^{-2 x}\) and \(y=e^{-x}+k\) intersect at two points. (3 marks)

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a. \(\text{See Worked Solutions}\)

b. \(-\dfrac{1}{4} < k < 0 \)

Show Worked Solution

a.	\(A\)	\(= \int_0^{\ln2} e^{-2x}-(e^{-x}-\dfrac{1}{4})\ dx\)
		\(=\Big{[}-\dfrac{1}{2} e^{-2x}+e^{-x}+\dfrac{1}{4}x \Big{]}_0^{\ln2} \)
		\(=\Big{[}(-\dfrac{1}{2} e^{-2\ln2}+e^{-\ln2}+\dfrac{1}{4}\ln2)-(-\dfrac{1}{2}e^0+e^0-0)\Big{]} \)
		\(=\Big{[}(-\dfrac{1}{2} e^{\ln{(2^{-2})}}+e^{\ln{(2^{-1})}}+\dfrac{1}{4}\ln2+\dfrac{1}{2}-1)\Big{]} \)
		\(=\Big{[}(-\dfrac{1}{2} e^{\ln \frac{1}{4}}+e^{\ln \frac{1}{2}}+\dfrac{1}{4}\ln2-\dfrac{1}{2}\Big{]} \)
		\(=-\dfrac{1}{2} \times \dfrac{1}{4} +\dfrac{1}{2}+\dfrac{1}{4}\ln2-\dfrac{1}{2} \)
		\(=\dfrac{1}{4}\ln2-\dfrac{1}{8} \)

b. \(\text{Intersection occurs when}\)

\(e^{-2x}\)	\(=e^{-x}+k\)
\(e^{-2x}-e^{-x}-k\)	\(=0\)

\(\text{Let}\ X=e^{-x} \)

\(X^2-X-k=0 \)

\(X\)	\(=\dfrac{1\pm \sqrt{1^2-4(1)(-k)}}{2} \)
	\(=\dfrac{1\pm \sqrt{1+4k}}{2} \)

\(\text{2 solutions}\ \Rightarrow\ \Delta >0 \)

\(1+4k>0\ \ \Rightarrow \ k>-\dfrac{1}{4} \)

\(\text{Since}\ X=e^{-x} >0:\)

\(\Rightarrow\ \text{Both real solutions to the quadratic MUST be positive.}\)

\(\sqrt{1+4k}\)	\(<1\)
\(1+4k\)	\(<1\)
\(k\)	\(<0\)

\(\therefore\ -\dfrac{1}{4} < k < 0 \)

♦♦♦ Mean mark (b) 14%.

Calculus, 2ADV C4 2023 HSC 5 MC

The diagram shows the graph `y=f(x)`, where `f(x)` is an odd function.

The shaded area is 1 square unit.

The number `a`, where `a > 1`, is chosen so that `int_0^a f(x)\ dx=0`.

What is the value of `int_{-a}^1 f(x)\ dx` ?

`-1`
`0`
`1`
`3`

Show Answers Only

`A`

Show Worked Solution

`text{Since}\ \ int_0^a f(x)\ dx=0\ and \ int_0^1 f(x)\ dx=-1\ \ text{(given)}`

`int_1^a f(x)\ dx=1`

`:. int_{-a}^{-1} f(x)\ dx=-1\ \ text{(f(x) is odd)}`

`:. int_{-a}^1 f(x)\ dx=-1`

`=>A`

♦ Mean mark 49%.

Calculus, 2ADV C4 2022 HSC 28

The graph of the circle `x^2+y^2=2` is shown.

The interval connecting the origin, `O`, and the point `(1,1)` makes an angle `theta` with the positive `x`-axis.

By considering the value of `theta`, find the exact area of the shaded region, as shown on the diagram. (2 marks)

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Part of the hyperbola `y=(a)/(b-x)-1` which passes through the points `(0,0)` and `(1,1)` is drawn with the circle `x^2+y^2=2` as shown.

Show that `a=b=2`. (2 marks)

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Using parts (a) and (b), find the exact area of the region bounded by the hyperbola, the positive `x`-axis and the circle as shown on the diagram. (3 marks)

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a. `(pi-2)/4\ text{u}^2`

b. `text{Proof (See Worked Solutions)}`

c. `(8ln2+pi-6)/4\ text{u}^2`

Show Worked Solution

a. `tan theta=1\ \ =>\ \ theta = pi/4`

`text{Using Pythagoras,}`

`r=sqrt(1^2+1^2)=sqrt2`

`text{Shaded Area}`	`=A_text{sector}-A_Delta`
	`=(pi/4)/(2pi) xx pi r^2-1/2 xx b xx h`
	`=1/8xxpixx(sqrt2)^2-1/2xx1xx1`
	`=pi/4-1/2`
	`=(pi-2)/4\ text{u}^2`

Mean mark (a) 54%.

b. `text{Show}\ \ a=b=2`

`y=(a)/(b-x)-1\ \ text{passes through}\ \ (0,0):`

`0`	`=a/(b-0)-1`
`a/b`	`=1`
`a`	`=b`

`y=(a)/(b-x)-1\ \ text{passes through}\ \ (1,1):`

`1`	`=a/(b-1)-1`
`a/(b-1)`	`=2`
`a`	`=2(b-1)`
`a`	`=2b-2`
`b`	`=2b-2\ \ text{(using}\ a=b)`
`b`	`=2`

`:.a=b=2`

♦ Mean mark (b) 47%.

c.	`int_0^1 2/(2-x)-1\ dx`	`=int_0^1-2 xx (-1)/(2-x)-1\ dx`
		`=[-2ln\|2-x\|-x]_0^1`
		`=[(-2ln1-1)-(-2ln2-0)]`
		`=-1+2ln2`

`:.\ text{Total Area}`	`=2ln2-1 + (pi-2)/4`
	`=(8ln2-4+pi-2)/4`
	`=(8ln2+pi-6)/4\ text{u}^2`

♦♦ Mean mark (c) 36%.

Calculus, 2ADV C4 2022 HSC 16

The parabola `y=x^2` meets the line `y=2 x+3` at the points `(-1,1)` and `(3,9)` as shown in the diagram.

Find the area enclosed by the parabola and the line. (3 marks)

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`32/3\ text{u}^2`

Show Worked Solution

`A`	`=int_(-1)^(3) 2x+3-x^2\ dx`
	`=[x^2+3x-1/3 x^3]_(-1)^3`
	`=(9+9-27/3)-(1-3+1/3)`
	`=9-(-5/3)`
	`=32/3\ text{u}^2`

Calculus, 2ADV C4 2022 HSC 8 MC

The graph of the even function `y=f(x)` is shown.

The area of the shaded region `A` is `1/2` and the area of the shaded region `B` is `3/2`.

What is the value of `int_(-2)^(2)f(x)\ dx`?

`4`
`2`
`-2`
`-4`

Show Answers Only

`C`

Show Worked Solution

`text{Areas under the}\ x text{-axis are negative}`

`int_0 ^2 f(x)\ dx = 1/2-3/2=-1`

`text{S}text{ince}\ \ f(x)\ \ text{is even:}`

`int_(-2) ^2 f(x)\ dx = 2int_0 ^2 f(x)\ dx = -2`

`=>C`

Mean mark 52%.

Calculus, 2ADV C4 2021 HSC 28

The region bounded by the graph of the function `f(x) = 8-2^x` and the coordinate axes is shown

Show that the exact area of the shaded region is given by `24-7/ln2`. (3 marks)

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A new function `g(x)` is found by taking the graph of `y =-f(-x)` and translating it by 5 units to the right.
Sketch the graph of `y = g(x)` showing the `x`-intercept and the asymptote. (2 marks)

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Hence, find the exact value of `int_2^5 g(x)\ dx`. (1 mark)

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a. `text(See Worked Solution)`

c. `7/(ln2)-24`

Show Worked Solution

a. `xtext(-intercept occurs when)`

`8-2^x = 0 \ => \ x = 3`

`text(Area)`	`= int_0^3 8-2^x\ dx`
	`= [8x-(2^x)/(ln2)]_0^3`
	`= 24-8/(ln 2)-(0-1/(ln2))`
	`= 24-8/(ln2) + 1/(ln2)`
	`= 24-7/(ln2)\ \ text(u²)`

♦ Mean mark part (b) 48%.

`y = f(-x) -> text(reflect)\ \ y = f(x)\ \ text(in the)\ ytext(-axis)`

`y =-f(-x) -> text(reflect)\ \ y = f(-x)\ \ text(in the)\ xtext(-axis)`

♦♦♦ Mean mark part (c) 13%.

c. `int_2^5 g(x)\ dx\ \ text{is the same area as found in part (a)}`

`text(except it is below the)\ xtext(-axis.)`

`:. int_2^5 g(x)\ dx = 7/(ln2)-24`

Calculus, 2ADV C4 2021 HSC 24

The curve `y = 3/(x - 1)` intersects the line `y = 3/2 x` at the point (2, 3).

The region bounded by the curve `y = 3/(x - 1)`, the line `y = 3/2 x`, the `x`-axis and the line `x = 4` is shaded in the diagram.

Find the exact area of the shaded region. (3 marks)

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`3 + 3 log_e 3\ text(u)²`

Show Worked Solution

`text(Area)`	`= int_0^2 3/2 x\ dx + int_2^4 3/(x-1)\ dx`
	`= [3/4 x^2]_0^2 + 3[log_e(x-1)]_2^4`
	`= 12/4 + 3[log_e 3-log_e 1]`
	`= 3 + 3 log_e 3\ \ text(u²)`

Calculus, 2ADV C4 2020 HSC 30

The diagram shows two parabolas `y = 4x-x^2` and `y = ax^2`, where `a > 0`. The two parabolas intersect at the origin, `O`, and at `A`.

Show that the `x`-coordinate of `A` is `4/(a + 1)`. (2 marks)

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Find the value of `a` such that the shaded area is `16/3`. (4 marks)

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a. `text(See Worked Solutions)`

b. `sqrt2-1`

Show Worked Solution

a. `text(Intersection occurs when)`

`4x-x^2`	`= ax^2`
`x^2(a + 1)-4x`	`= 0`
`x[x(a + 1)-4]`	`= 0`

`x(a+1)-4`	`=0\ \ \ text(or)`	`x=0`
`:. x_A`	`=4/(a + 1)`

♦ Mean mark part (b) 48%.

b.	`text(Area)`	`= int_0^(4/(a + 1)) 4x-x^2\ dx-int_0^(4/(a + 1)) ax^2\ dx`
	`16/3`	`= int_0^(4/(a + 1)) 4x-(1 + a)x^2\ dx`
	`16/3`	`= [2x^2-((1 + a)/3) x^3]_0^(4/(a + 1))`
	`16/3`	`= 2(4/(a + 1))^2-((1 + a)/3)(4/(a + 1))^3`
	`16/3`	`= 32/((a + 1)^2)-64/3 · 1/((a + 1)^2)`
	`16`	`= 96/((a + 1)^2)-64/((a + 1)^2)`

`16(a + 1)^2`	`= 32`
`(a + 1)^2`	`= 2`
`a + 1`	`= sqrt2,\ \ \ (a > 0)`
`:. a`	`= sqrt2-1`

Calculus, 2ADV C4 2020 HSC 7 MC

The diagram show the graph `y = f(x)`, which is made up of line segments and a semicircle.

What is the value of `int_0^12 f(x)\ dx`?

`24 + 2pi`
`24 + 4pi`
`30 + 2pi`
`30 + 4pi`

Show Answers Only

`A`

Show Worked Solution

`text(Consider the interval between)\ \ x=8 and x=12:`

`text(Area above and below the)\ xtext(-axis are equal.)`

`int_8^12 f(x) = 0`

♦ Mean mark 48%.

`:. int_0^12 f(x)\ dx`	`= int_0^8 f(x)\ dx`
	`=\ text(Area of rectangle + area of semi-circle)`
	`= 8 xx 3 + 1/2 xx pi xx 2^2`
	`= 24 + 2pi`

`=>A`

Calculus, 2ADV C4 2019 MET-N 15 MC

The area bounded by the graph of `y = f(x)`, the line `x = 2`, the line `x = 8` and the `x`-axis, as shaded in the diagram below, is `3log_e(13)`

The value of `int_4^10 3 f(x-2)\ dx` is

`3log_e(13)`
`9log_e(13)`
`6log_e(39)`
`9log_e(11)`

Show Answers Only

`B`

Show Worked Solution

`A_1 = int_2^8 f(x)\ dx = 3 log_e 13`

`f(x-2) = f(x) \ text(shifted 2 units to right.)`

`int_2^8 f(x)\ dx = int_4^10 f(x-2)\ dx`

`:. \ 3 int_4^10 f(x-2)\ dx`	`= 3 xx 3log_e 13`
	`= 9 log_e 13`

`=> \ B`

Calculus, 2ADV C4 2019 MET1 7

The shaded region in the diagram below is bounded by the vertical axis, the graph of the function with rule `f(x) = sin(pix)` and the horizontal line segment that meets the graph at `x = a`, where `1 <= a <= 3/2`.

Let `A(a)` be the area of the shaded region.

Show that `A(a) = 1/pi-1/pi cos(a pi)-a sin (a pi)`. (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text{Strategy 1}`

`text(Consider the areas above:)`

`int_0^a \sin (pi x)`	`\ = text(Area 1 – Area 3)`
	`=[-{1}/{pi} cos (pi x)]_0^a`
	`=-{1}/{pi} cos (a pi)-(-{1}/{pi})`
	`=1/pi-1/picos(a pi)`

`text(Area 2 + Area 3 (rectangle) )`

`=-sin(a pi) \times a`

`=-a sin (a pi) \ text{(Area must be +ve)}`

`\therefore \ text(Shaded area)`	`\ =text(Area 1 + Area 2)`
	`=1/pi-1/pi cos(a pi)-a sin(a pi)`

`text{Strategy 2}`

`text(Lower border of shaded area:)\ y = f(a) = sin(a pi)`

`text(Area)`	`= int_0^a sin(pi x)-sin (a pi)\ dx`
	`= [-1/pi cos (pi x)-x sin(a pi)]_0^a`
	`= [-1/pi cos(pi a)-a sin (a pi)-(-1/pi-0)]`
	`= 1/pi-1/pi cos (a pi)-a sin(a pi)`

Calculus, 2ADV C4 2019 HSC 16c

The diagram shows the region `R`, bounded by the curve `y = x^r`, where `r >= 1`, the `x`-axis and the tangent to the curve at the point `(1, 1)`.

Show that the tangent to the curve at `(1, 1)` meets the `x`-axis at

`qquad ((r-1)/r, 0)`. (2 marks)

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Using the result of part (i), or otherwise, show that the area of the region `R` is

`qquad (r-1)/(2r (r + 1))`. (2 marks)

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Find the exact value of `r` for which the area of `R` is a maximum. (3 marks)

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i. `text(Proof)\ text{(See Worked Solutions)}`

ii. `text(Proof)\ text{(See Worked Solutions)}`

iii. `r = 1 + sqrt 2`

Show Worked Solution

i.	`y`	`= x^r`
	`(dy)/(dx)`	`= r x^(r-1)`

`text(When)\ \ x = 1, \ (dy)/(dx) = r`

♦♦ Mean mark part (i) 31%.

`text(Equation of tangent:)`

`y-1`	`= r(x-1)`
`y`	`= rx-r + 1`

`text(When)\ \ y = 0:`

`rx-r + 1`	`= 0`
`rx`	`= r-1`
`x`	`= (r-1)/r`

`:.\ text(T)text(angent meets)\ x text(-axis at)\ \ ((r-1)/r, 0)`

ii. `text(Area under curve)`

♦♦♦ Mean mark part (ii) 21%.

`= int_0^1 x^r`

`= [1/(r + 1) ⋅ x^(r + 1)]_0^1`

`= 1/(r + 1) xx 1^(r + 1)-0`

`= 1/(r + 1)`

`text(Area under tangent)`

`= 1/2 xx b xx h`

`= 1/2 (1-(r-1)/r) xx 1`

`= 1/2 (1-(r-1)/r)`

`:. R`	`= 1/(r + 1)-1/2(1-(r-1)/r)`
	`= 1/(r + 1)-1/(2r) [r-(r-1)]`
	`= 1/(r + 1)-1/(2r)`
	`= (2r-(r + 1))/(2r(r + 1))`
	`= (r-1)/(2r(r + 1))`

iii.	`R`	`= (r-1)/(2r(r + 1)) = (r-1)/(2r^2 + 2r)`
	`(dR)/(dr)`	`= ((2r^2 + 2r) xx 1-(r-1)(4r + 2))/(2r^2 + 2r)^2`
		`= (2r^2 + 2r-4r^2-2r + 4r + 2)/(2r^2 + 2r)^2`
		`= (-2r^2 + 4r + 2)/(2r^2 + 2r)^2`
		`= (-2(r^2-2r-1))/(2r^2 + 2r)^2`

`text(Find)\ \ r\ \ text(when)\ \ (dR)/(dr) = 0:`

♦♦ Mean mark part (iii) 23%.

`r^2-2r-1 = 0`

`r`	`= (2 +- sqrt((-2)^2-4 xx 1 xx (-1)))/2 `
	`= (2 +- sqrt 8)/2`
	`= 1 + sqrt 2\ \ (r >= 1)`

`qquadr qquad`	`qquad 1 qquad`	`\ \ 1 + sqrt 2\ \ `	`qquad 3 qquad`
`(dR)/(dr)`	`1/4`	`0`	`-1/144`

`:. R_text(max)\ text(occurs when)\ \ r = 1 + sqrt 2`

Calculus, 2ADV C4 2019 HSC 12d

The diagram shows the graph of `y = (3x)/(x^2 + 1)`.

The region enclosed by the graph, the `x`-axis and the line `x = 3` is shaded.

Calculate the exact value of the area of the shaded region. (3 marks)

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`3/2 ln 10\ text(u²)`

Show Worked Solution

`text(Area)`	`= int_0^3 (3x)/(x^2 + 1)\ dx`
	`= 3/2 int_0^3 (2x)/(x^2 + 1)\ dx`
	`= 3/2[ln(x^2 + 1)]_0^3`
	`= 3/2(ln 10-ln 1)`
	`= 3/2 ln10\ \ text(u²)`

Calculus, 2ADV C4 2007* HSC 10a

An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for `t >= 6`.

The object is initially at the origin. During which time(s) is the displacement of the object decreasing? (1 mark)

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If the object travels 7 units in the first 4 seconds, estimate the time at which the object returns to the origin. Justify your answer. (2 marks)

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Sketch the displacement, `x`, as a function of time. (2 marks)

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a. `t > 5\ \ text(seconds)`

b. `7.4\ \ text(seconds)`

Show Worked Solution

a. `text(Displacement is reducing when the velocity is negative.)`

`:. t > 5\ \ text(seconds)`

b. `text(At)\ B,\ text(the displacement) = 7\ text(units)`

`text(Consider the displacement from)\ B\ text(to)\ D:`

`text(Since the area below the graph from)\ B\ text(to)\ C\ text(equals )`

`text(the area above the graph from)\ C\ text(to)\ D,\ text(there is no )`

`text(change in displacement from)\ B\ text(to)\ D.`

`text(Consider)\ t >= 6,`

`text(Time required to return to origin:)`

`t=d/v= 7/5= 1.4\ \ text(seconds)`

`:.\ text(The particle returns to the origin after 7.4 seconds.)`

Calculus, 2ADV C4 2018* HSC 15c

The shaded region is enclosed by the curve `y = x^3-7x` and the line `y = 2x`, as shown in the diagram. The line `y = 2x` meets the curve `y = x^3-7x` at `O(0, 0)` and `A(3, 6)`. Do NOT prove this.

Use integration to find the area of the shaded region. (2 marks)

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Use the Trapezoidal rule and four function values to approximate the area of the shaded region. (2 marks)

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The point `P` is chosen on the curve `y = x^3-7x` so that the tangent at `P` is parallel to the line `y = 2x` and the `x`-coordinate of `P` is positive

Show that the coordinates of `P` are `(sqrt 3, -4 sqrt 3)`. (2 marks)

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Using the perpendicular distance formula `|ax_1 + by_1 + c|/sqrt(a^2 + b^2)`, find the area of `Delta OAP`. (2 marks)

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i. `81/4\ text(units)^2`

ii. `18\ text(u)^2`

iii. `text(Proof)\ \ text{(See Worked Solutions)}`

iv. `9 sqrt 3\ text(units)^2`

Show Worked Solution

i.	`text(Area)`	`= int_0^3 2x-(x^3-7x)\ dx`
		`= int_0^3 9x-x^3\ dx`
		`= [9/2 x^2-1/4 x^4]_0^3`
		`= [(9/2 xx 3^2-1/4 xx 3^4)-0]`
		`= 81/2-81/4`
		`= 81/4\ text(units)^2`

ii. `f(x) = 9x-x^3`

`text(Area)~~ 1/2[0 + 2(8 + 10) + 0]~~ 1/2(36)~~ 18\ text(u)^2`

iii. `y = x^3-7x`

`(dy)/(dx) = 3x^2-7`

`text(Find)\ x\ text(such that)\ \ (dy)/(dx) = 2:`

`3x^2-7`	`= 2`
`3x^2`	`= 9`
`x^2`	`= 3`
`x`	`= sqrt 3 qquad (x > 0)`

`y= (sqrt 3)^3-7 sqrt 3= 3 sqrt 3-7 sqrt 3= -4 sqrt 3`

`:. P\ \ text(has coordinates)\ (sqrt 3, -4 sqrt 3)`

iv.

`text(dist)\ OA= sqrt((3-0)^2 + (6-0)^2)= sqrt 45= 3 sqrt 5`

`text(Find)\ _|_\ text(distance of)\ P\ text(from)\ OA:`

`P(sqrt 3, -4 sqrt 3),\ \ 2x-y=0`

`_|_\ text(dist)= |(2 sqrt 3 + 4 sqrt 3)/sqrt (3 + 2)|= (6 sqrt 3)/sqrt 5`

`:.\ text(Area)= 1/2 xx 3 sqrt 5 xx (6 sqrt 3)/sqrt 5= 9 sqrt 3\ text(units)^2`

Calculus, 2ADV C4 2018 HSC 15b

The diagram shows the region bounded by the curve `y = 1/(x + 3)` and the lines `x = 0`, `x = 45` and `y = 0`. The region is divided into two parts of equal area by the line `x = k`, where `k` is a positive integer.

What is the value of the integer `k`, given that the two parts have equal areas? (3 marks)

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`9`

Show Worked Solution

`text(Total Area)`	`= int_0^45 1/(x + 3)`
	`= [ln (x + 3)]_0^45`
	`= ln 48-ln 3`
	`= ln 16`

`=> int_0^k 1/(x + 3)`	`= 1/2 xx ln 16`
`[ln (x + 3)]_0^k`	`= ln 16^(1/2)`
`ln (k + 3)-ln 3`	`= ln 4`
`ln ((k + 3)/3)`	`= ln 4`
`(k + 3)/3`	`= 4`
`:. k`	`= 4 xx 3-3`
	`= 9`

Calculus, 2ADV C4 2018 HSC 10 MC

A trigonometric function `f(x)` satisfies the condition

`int_0^pi f(x)\ dx != int_pi^(2pi) f(x)\ dx.`

Which function could be `f(x)`?

`f(x) = sin (2x)`
`f(x) = cos (2x)`
`f(x) = sin (x/2)`
`f(x) = cos (x/2)`

Show Answers Only

`D`

Show Worked Solution

`text(Consider options A and C)`

`text(Consider options B and D)`

`text(When)\ \ y = cos\ x/2 ,`

`int_0^pi f(x)\ dx != int_pi^(2 pi) f(x)\ dx`

`=> D`

Calculus, 2ADV C4 2018 HSC 7 MC

The diagram shows the graph of `y = f(x)` with intercepts at `x = -1, 0, 3 and 4.`

The area of shaded region `R_1` is 2.

The area of shaded region `R_2` is 3.

It is given that `int_0^4 f(x)\ dx = 10`.

What is the value of `int_(-1)^3 f(x)\ dx?`

Show Answers Only

`C`

Show Worked Solution

`int_0^4 f(x)\ dx = 10`

♦ Mean mark 36%.

`:. R_3-R_2`	`= 10`
`R_3`	`= 13`

`int_(-1)^3 f(x)\ dx`	`= R_3-R_1`
	`= 13-2=11`

`=> C`

Calculus, 2ADV C4 EQ-Bank 28

Let `f(x) = 2e^(-x/5)\ \ \ text(for)\ \ x>=0`

A right-angled triangle `OQP` has vertex `O` at the origin, vertex `Q` on the `x`-axis and vertex `P` on the graph of `f(x)`, as shown. The coordinates of `P` are `(x, f(x)).`

Find the area, `A`, of the triangle `OPQ` in terms of `x`. (1 mark)

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Find the maximum area of triangle `OQP` and the value of `x` for which the maximum occurs. (3 marks)

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Let `S` be the point on the graph of `f(x)` on the `y`-axis and let `T` be the point on the graph of `f(x)` with the `y`-coordinate `1/2`. Find the area of the region bounded by the graph of `f(x)` and the line segment `ST`. (2 marks)

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a. `x e^(-x/5)`

b. `5/e\ text(u)^2`

c. `25/4 log_e (4)-15/2\ text(u²)`

Show Worked Solution

a. `text(Area)= 1/2 xx b xx h= 1/2x(2e^(-x/5))= xe^(-x/5)`

b. `text(Stationary point when)\ \ (dA)/(dx) = 0:`

♦ Mean mark (b) (Vic) 35%.

`x(-1/5 e^(-x/5)) + e^(-x/5)`	`= 0`
`e^(-x/5)(1-x/5)`	`= 0`

`:. x= 5\ \ \ \ (e^(-x/5) >0,\ \ text(for all)\ x)`

`text(When)\ \ x = 5,\ \ A= xe^(-x/5)= 5e^-1`

`:. A_max = 5/e\ text(u², when)\ \ x = 5`

c. `text(Find)\ S:`

`F(0) = 2\ \ =>\ \ S(0, 2)`

♦♦ Mean mark (c) (Vic) 32%.

`text(Find)\ T:\ \ \ `	`2e^(-x/5)`	`= 1/2`
	`e^(-x/5)`	`= 1/4`
	`-x/5`	`= log_e (1/4)`
	`x`	`= 5 log_e (4)`

`=> T(5log_e(4), 1/2)`

`:.\ text(Area)`	`= text(Area)\ SOAT-int_0^(5 log_e(4)) (2e^(-x/5)) dx`
	`=1/2h(a+b) + 10 [e^(-x/5)]_0^(5 log_e (4))`
	`= 5/2 log_e (4) (2 + 1/2) + 10 [e^(-log_e (4))-e^0]`
	`= 25/4 log_e (4) +10 (1/4-1)`
	`= 25/4 log_e (4)-15/2\ text(u)^2`

Calculus, 2ADV C3 EQ-Bank 34

The graph of `f(x) = sqrt x (1-x)` for `0<=x<=1` is shown below.

Calculate the area between the graph of `f(x)` and the `x`-axis. (2 marks)

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For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of `f(x)` is `(1-3x)/(2 sqrt x)`. (1 mark)

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The edges of the right-angled triangle `ABC` are the line segments `AC` and `BC`, which are tangent to the graph of `f(x)`, and the line segment `AB`, which is part of the horizontal axis, as shown below.

Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis.

Find the equation of the line through `B` and `C` in the form `y = mx + c`, for `theta = 45^@`. (3 marks)

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a. `4/15\ text(units)^2`

b. `text(Proof)\ \ text{(See Worked Solutions)}`

c. `y = -x + 1`

Show Worked Solution

a.	`text(Area)`	`= int_0^1 (sqrt x-x sqrt x)\ dx`
		`= int_0^1 (x^(1/2)-x^(3/2))\ dx`
		`= [2/3 x^(3/2)-2/5 x^(5/2)]_0^1`
		`= (2/3-2/5)-(0-0)`
		`= 10/15-6/15`
		`= 4/15\ text(units)^2`

b.	`f (x)`	`= x^(1/2)-x^(3/2)`
	`f^{prime}(x)`	`= 1/2 x^(-1/2)-3/2 x^(1/2)`
		`= 1/(2 sqrt x)-(3 sqrt x)/2`
		`= (1-3x)/(2 sqrt x)\ \ text(.. as required.)`

c. `m_(AC) = tan 45^@=1`

♦♦♦ Mean mark (Vic) part (c) 20%.
MARKER’S COMMENT: Most successful answers introduced a pronumeral such as `a=sqrtx` to solve.

`=> m_(BC) =-1\ \ (m_text(BC) _|_ m_(AC))`

`text(At point of tangency of)\ BC,\ f^{prime}(x) =-1`

`(1-3x)/(2 sqrt x)`	`=-1`
`1-3x`	`=-2sqrtx`
`3x-2sqrt x-1`	`=0`

`text(Let)\ \ a=sqrtx,`

`3a^2-2a-1`	`=0`
`(3a+1)(a-1)`	`=0`
`a=1 or -1/3`

`:. sqrt x`	`=1`	`or`	`sqrt x=-1/3\ \ text{(no solution)}`
`x`	`=1`

`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`

`text(Equation of)\ \ BC, \ m=-1, text{through (1,0):}`

`y-0`	`=-1(x-1)`
`y`	`=-x+1`

Calculus, 2ADV C4 2017 HSC 14d

The shaded region shown is enclosed by two parabolas, each with `x`-intercepts at `x = –1` and `x = 1`.

The parabolas have equations `y = 2k (x^2 - 1)` and `y = k(1 - x^2)`, where `k > 0`.

Given that the area of the shaded region is 8, find the value of `k`. (3 marks)

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`k = 2`

Show Worked Solution

`text(Area)`	`= int_(-1)^1 k (1-x^2) dx-int_(-1)^1 2k (x^2-1) dx`
`8`	`= 2 int_0^1 k-kx^2-2kx^2 + 2k\ dx`
`8`	`= 2 int_0^1 3k-3kx^2\ dx`
`8`	`= 2 [3kx-kx^3]_0^1`
`8`	`= 2 [(3k-k)-0]`
`8`	`= 4k`
`:. k`	`= 2`

Calculus, 2ADV C4 2016 HSC 13d

The curve `y = sqrt 2 cos (pi/4 x)` meets the line `y = x` at `P(1, 1)`, as shown in the diagram.

Find the exact value of the shaded area. (3 marks)

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`(4/pi-1/2)\ text(u²)`

Show Worked Solution

`text(Shaded Area)`

`= int_0^1 sqrt 2 cos (pi/4 x)\ dx-int_0^1 x\ dx`

`= int_0^1 (sqrt 2 cos (pi/4 x)-x)\ dx`

`= [sqrt 2 xx 4/pi sin (pi/4 x)-x^2/2]_0^1`

`= [((4 sqrt 2)/pi sin\ pi/4-1/2)-0]`

`= (4 sqrt 2)/pi xx 1/sqrt 2-1/2`

`= (4/pi-1/2)\ text(u²)`

Calculus, 2ADV C4 2016 HSC 9 MC

What is the value of `int_-3^2 |\ x + 1\ |\ dx?`

`5/2`
`11/2`
`13/2`
`17/2`

Show Answers Only

`C`

Show Worked Solution

♦♦♦ Mean mark 19%.

`int_-3^2 \|\ x + 1\ \|\ dx`	`=\ text(Area of 2 triangles)`
	`= 1/2 xx 2 xx 2 + 1/2 xx 3 xx 3`
	`= 13/2`

`=> C`

Calculus, 2ADV C4 EQ-Bank 4 MC

The diagram below is the graph of `y = x^2-x-6`

What is the correct expression for the area bounded by the `x`-axis and the graph `y = x^2-x-6` between `0 <= x <= 4`?

`A = int_0^4 x^2-x-6\ dx`
`A = |\ int_0^3 x^2-x-6\ dx\ | + int_3^4 x^2-x-6\ dx`
`A = int_0^3 x^2-x-6\ dx + |\ int_3^4 x^2-x-6\ dx|`
`A = |\ int_0^4 x^2-x-6\ dx\ |`

Show Answers Only

`B`

Show Worked Solution

`y= x^2-x-6= (x-3)(x + 2)`

`text(Graph intersects the)\ xtext(-axis at)\ (3,0)`

`:. A = |\ int_0^3 x^2-x-6\ dx\ | + int_3^4 x^2-x-6\ dx`

`=> B`

Calculus, 2ADV C4 2007 HSC 7b

The diagram shows the graphs of `y = sqrt 3 cos x` and `y = sin x`. The first two points of intersection to the right of the `y`-axis are labelled `A` and `B`.

Solve the equation `sqrt 3 cos x = sin x` to find the `x`-coordinates of `A` and `B`. (2 marks)

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Find the area of the shaded region in the diagram. (3 marks)

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a. `pi/3, (4 pi)/3`

b. `4\ text(u²)`

Show Worked Solution

a. `sqrt 3 cos x`	`= sin x`
`tan x`	`= sqrt 3`

`text(S)text(ince)\ tan\ pi/3 = sqrt 3 and tan\ text(is)`

`text(positive in 1st/3rd quadrants:)`

`x`	`= pi/3 , pi + pi/3`
	`= pi/3, (4 pi)/3`

`:. text(The)\ x text(-coordinates of)\ A and B`

`text(are)\ \ x = pi/3, (4 pi)/3.`

b. `text(Shaded Area)`

`= int_(pi/3)^((4 pi)/3) sin\ x\ dx-int_(pi/3)^((4 pi)/3) sqrt 3\ cos\ x\ dx`

`= int_(pi/3)^((4 pi)/3) sin\ x-sqrt 3\ cos\ x\ dx`

`= [-cos\ x- sqrt 3\ sin\ x]_(pi/3)^((4 pi)/3)`

`= [(-cos\ (4 pi)/3-sqrt 3\ sin\ (4 pi)/3)-(-cos\ pi/3-sqrt 3\ sin\ pi/3)]`

`= [(1/2 + sqrt 3 xx (sqrt 3)/2)-(-1/2-sqrt 3 xx (sqrt 3)/2)]`

`= [2-(-2)]= 4\ text(u²)`

Calculus, 2ADV C4 2015 HSC 16a

The diagram shows the curve with equation `y = x^2-7x + 10`. The curve intersects the `x`-axis at points `A and B`. The point `C` on the curve has the same `y`-coordinate as the `y`-intercept of the curve.

Find the `x`-coordinates of points `A and B`. (1 mark)

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Write down the coordinates of `C`. (1 mark)

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Evaluate `int _0^2 (x^2-7x + 10)\ dx`. (1 mark)

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Hence, or otherwise, find the area of the shaded region. (2 marks)

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a. `A = 2,\ \ B = 5`

b. `(7, 10)`

c. `8 2/3`

d. `16 1/3\ text(u²)`

Show Worked Solution

a. `y`	`= x^2-7x + 10`
	`= (x-2) (x-5)`

`:.x = 2 or 5`

`:.\ \ x text(-coordinate of)\ \ A = 2`

`x text(-coordinate of)\ \ B = 5`

b. `y\ text(intercept occurs when)\ \ x = 0`

`=>y text(-intercept) = 10`

`C\ text(occurs at intercept:)`

`y`	`= x^2-7x + 10`	`\ \ \ \ \ text{… (1)}`
`y`	`= 10`	`\ \ \ \ \ text{… (2)}`

`(1) = (2)`

`x^2-7x + 10`	`= 10`
`x^2-7x`	`= 10`
`x (x-7)`	`= 10`

`x = 0 or 7`

`:.\ C\ \ text(is)\ \ (7, 10)`

c. `int_0^2 (x^2-7x + 10)\ dx`

`= [1/3 x^3-7/2 x^2 + 10x]_0^2`

`= [(1/3 xx 2^3-7/2 xx 2^2 + 10 xx 2)-0]`

`= 8/3-14 + 20`

`= 8 2/3`

`A_1 = A_2`

♦ Mean mark (d) 49%.

`A_2 = 8 2/3\ text(u²)\ \ \ \ text{(from part (c))}`

`text(Let)\ \ D\ \ text(be)\ \ (7, 0)`

`text(Shaded Area)`

`= text(Area of)\ \ Delta ACD-A_2`

`= 1/2 bh-8 2/3`

`= 1/2 xx 5 xx 10-8 2/3`

`= 16 1/3\ text(u²)`

Calculus, 2ADV C4 2015 HSC 10 MC

The diagram shows the area under the curve `y = 2/x` from `x = 1` to `x = d`.

What value of `d` makes the shaded area equal to `2`?

`e`
`e + 1`
`2e`
`e^2`

Show Answers Only

`A`

Show Worked Solution

`int_1^d 2/x\ dx = 2`

`[2 log_e x]_1^d = 2`

`2 log_e d-2 log_e 1`	`= 2`
`2 log_e d`	`= 2`
`log_e d`	`= 1`
`:. d`	`= e`

`=> A`

Calculus, 2ADV C4 2015 HSC 7 MC

The diagram shows the parabola `y = 4x-x^2` meeting the line `y = 2x` at `(0, 0)` and `(2, 4)`.

Which expression gives the area of the shaded region bounded by the parabola and the line?

`int_0^2 x^2-2x\ dx`
`int_0^2 2x-x^2\ dx`
`int_0^4 x^2-2x\ dx`
`int_0^4 2x-x^2\ dx`

Show Answers Only

`B`

Show Worked Solution

`text(Shaded Area)`	`= int_0^2 4x-x^2\ dx-int_0^2 2x\ dx`
	`= int_0^2 4x-x^2-2x\ dx`
	`= int_0^2 2x-x^2\ dx`

`=> B`

Calculus, 2ADV C4 2006 HSC 5b

Show that `d/dx log_e (cos x) = -tan x`. (1 mark)

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The shaded region in the diagram is bounded by the curve `y =tan x` and the lines `y =x` and `x = pi/4.`

Using the result of part (i), or otherwise, find the area of the shaded region. (3 marks)

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Show Answers Only

a. `text(Proof)\ \ text{(See Worked Solutions)}`

b. `(1/2 log_e 2-pi^2/32)\ text(u²)`

Show Worked Solution

a. `d/dx log_e (cos x)`	`= (-sin x)/cos x`
	`=-tan x\ …\ text(as required)`

b. `text(Shaded Area)`

`= int_0^(pi/4) tan x\ dx-int_0^(pi/4) x\ dx`

`= int_0^(pi/4) tan x-x\ dx`

`= [-log_e (cos x)-1/2 x^2]_0^(pi/4)`

`=[(-log_e (cos pi/4)-1/2 xx (pi^2)/16)-(-log_e(cos 0)-0)]`

`= -log_e 1/sqrt 2-pi^2/32 + log_e1`

`= -log_e 2^(-1/2)-pi^2/32`

`= (1/2 log_e 2-pi^2/32)\ text(u²)`

Calculus, 2ADV C4 2005 HSC 8b

The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola `y = x^2-3x + 2`, and the `x`-axis.

By considering the difference of two areas, find the area of the shaded region. (3 marks)

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Show Answers Only

`(pi-5/6)\ \ text(u²)`

Show Worked Solution

`text(Shaded Area = Area in the quarter circle less)`

`text(the area below the parabola between)\ x= 0 and 1.`

`text(Area of)\ 1/4\ text(circle)`	`= 1/4 xx pir^2`
	`= 1/4 xx pi xx 2^2`
	`= pi\ \ \ text(u²)`

`text(Area below the parabola between)\ x= 0 and 1`

`=int_0^1y\ dx`

`= int_0^1x^2-3x + 2\ dx`

`= [x^3/3-3/2x^2 + 2x]_0^1`

`= [(1/3-3/2 + 2)-0]`

`= 5/6`

`:.\ text(Shaded Area) = (pi-5/6)\ \ \ text(u²)`

Trigonometry, 2ADV T3 2006 HSC 7b

A function `f(x)` is defined by `f(x) = 1 + 2 cos x`.

Show that the graph of `y = f(x)` cuts the `x`-axis at `x = (2 pi)/3`. (1 mark)

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Sketch the graph of `y = f(x)` for `-pi <= x <= pi` showing where the graph cuts each of the axes. (3 marks)

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Find the area under the curve `y = f(x)` between `x = -pi/2` and `x = (2 pi)/3`. (3 marks)

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Show Answers Only

a. `text(Proof)\ \ text{(See Worked Solutions)}`

c. `((7 pi)/6 + sqrt 3 + 1)\ text(u²)`

Show Worked Solution

a. `f(x) = 1 + 2 cos x`

`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`

`1 + 2 cos x`	`= 0`
`2 cos x`	`=-1`
`cos x`	`=-1/2`

`:. x = (2 pi)/3\ …\ text(as required)`

c. `text(Area)`	`= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx`
	`= [x + 2 sin x]_(-pi/2)^((2 pi)/3)`
	`= [((2 pi)/3 + 2 sin (2 pi)/3)-((-pi)/2 + 2 sin (-pi)/2)]`
	`= ((2 pi)/3 + 2 xx sqrt 3/2)-((-pi)/2 +2(- 1))`
	`= (2 pi)/3 + sqrt(3) + pi/2 + 2`
	`= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)`

Calculus, 2ADV C4 2008 HSC 10a

In the diagram, the shaded region is bounded by `y = log_e (x-2)`, the `x`-axis and the line `x = 7`.

Find the exact value of the area of the shaded region. (5 marks)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

`5 log_e 5-4\ \ \ text(u²)`

Show Worked Solution

`text(Shaded Area)\ text{(} A_1 text{)}`	`=\ text(Rectangle)-A_2`
`text(Area of Rectangle)\ `	`= 7 xx log_e 5`

`text(Finding the Area of)\ \ A_2`

`y`	`= log_e (x-2)`
`x-2`	`= e^y`
`x`	`= e^y + 2`

`:. A_2`	`= int_0^(log_e 5) x\ dy`
	`= int_0^(log_e 5) e^y + 2\ dy`
	`= [e^y + 2y]_0^(log_e 5)`
	`= [(e^(log_e 5) + 2 log_e 5)-(e^0 + 0)]`
	`= (5 + 2 log_e 5)-1`
	`= 4 + 2 log_e 5`

`:.\ A_1`	`= 7 log_e 5-(4 + 2 log_e 5)`
	`= 5 log_e 5-4\ \ \ text(u²)`

Calculus, 2ADV C4 2014 HSC 12d

The parabola `y = −2x^2 + 8x` and the line `y = 2x` intersect at the origin and at the point `A`.

Find the `x`-coordinate of the point `A`. (1 mark)

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Calculate the area enclosed by the parabola and the line. (3 marks)

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Show Answers Only

a. `x=3`

b. `9\ text(u²)`

Show Worked Solution

`text(Need to find)\ x\ text(-co-ord of)\ A`

`y`	`= 2x\ \ \ \ \ …\ text{(i)}`
`y`	`= -2x^2 + 8x\ \ \ \ \ …\ text{(ii)}`

`text(Subst)\ y = 2x\ text(from)\ text{(i)}\ text(into)\ text{(ii)}`

`-2x^2 + 8x`	`= 2x`
`-2x^2 + 6x`	`= 0`
`-2x (x\ – 3)`	`= 0`

`:.\ x = 0\ text(or)\ 3`

`:.\ x\ text(-coordinate of)\ A\ text(is 3)`

b.	`text(Area)`	`= int_0^3 (-2x^2 + 8x)\ dx-int_0^3 2x\ dx`
		`= int_0^3 (-2x^2 + 6x)\ dx`
		`= [-2/3x^3 + 3x^2]_0^3`
		`= [(-2/3(3^3) + 3(3^2))-(0 + 0)]`
		`=-18 + 27`
		`= 9\ text(u²)`

Calculus, 2ADV C3 2009 HSC 10

`text(Let)\ \ f(x) = x-(x^2)/2 + (x^3)/3`

Show that the graph of `y = f(x)` has no turning points. (2 marks)

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Find the point of inflection of `y = f(x)`. (1 mark)

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i. Show that `1-x + x^2-1/(1 + x) = (x^3)/(1 + x)` for `x !=-1`. (1 mark)

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ii. Let `g(x) = ln (1 + x)`.

Use the result in part c.i. to show that `f^{prime} (x) >= g ^{prime}(x)` for all `x >= 0`. (2 marks)

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Sketch the graphs of `y = f(x)` and `y = g(x)` for `x >= 0`. (2 marks)

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Show that `d/(dx) [(1 + x) ln (1 + x)-(1 + x)] = ln (1 + x)`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Find the area enclosed by the graphs of `y = f(x)` and `y = g(x)`, and the straight line `x = 1`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `text{Proof (See Worked Solutions)}`

b. `(1/2, 5/12)`

c.i. `text{Proof (See Worked Solutions)}`

c.ii. `text{Proof (See Worked Solutions)}`

e. `text{Proof (See Worked Solutions)}`

f. `1 5/12-2ln2\ \ text(u²)`

Show Worked Solution

a.	`f(x) = x\ – (x^2)/2 + (x^3)/3`

♦♦ Mean mark 28% for all of Q10 (note that data for each question part is not available).

`text(Turning points when)\ f^{prime} (x) = 0`

`f^{prime}(x) = 1-x + x^2`

`x^2-x + 1 = 0`

`text(S)text(ince)\ \ Delta`	`= b^2-4ac`
	`= (-1)^2-4 xx 1 xx 1`
	`= -3 < 0 => text(No solution)`

`:.\ f(x)\ text(has no turning points)`

b.	`text(P.I. when)\ f^{prime prime}(x) = 0`

`f^{prime prime}(x)`	`=-1 + 2x = 0`
`2x`	`= 1`
`x`	`= 1/2`

`text(Check for change in concavity)`

`f^{prime prime}(1/4)`	`=-1/2 < 0`
`f^{prime prime}(3/4)`	`= 1/2 > 0`

`=>\ text(Change in concavity)`

`:.\ text(P.I. at)\ \ x = 1/2`

`f(1/2)`	`= 1/2-((1/2)^2)/2 + ((1/2)^3)/3`
	`= 1/2-1/8 + 1/24`
	`= 5/12`

`:.\ text(Point of Inflection at)\ (1/2, 5/12)`

c.i.

`text(Show)\ 1- x + x^2-1/(1 + x) = (x^3)/(1 + x),\ \ \ x !=-1`

`text(LHS)`	`= (1+x)/(1+x)-(x(1+x))/(1+x) + (x^2(1+x))/((1+x))-1/(1+x)`
	`= (1 + x-x-x^2 + x^2 + x^3-1)/(1+x)`
	`= (x^3)/(1+x)\ \ \ text(… as required)`

c.ii.	`text(Let)\ g(x) = ln(1+x)`
	`g^{prime} (x) = 1/(1 + x)`

`f^{prime} (x)-g^{prime} (x)`	`= 1-x + x^2-1/(1+x)`
	`= (x^3)/(1 + x)\ \ text{(using part (i))}`

`text(S)text(ince)\ (x^3)/(1 + x) >= 0\ text(for)\ x >= 0`

`f^{prime}(x)-g^{prime}(x) >= 0`

`f^{prime}(x) >= g^{prime}(x)\ text(for)\ x >= 0`

MARKER’S COMMENT: When 2 graphs are drawn on the same set of axes, you must label them.

e.	`text(Show)\ d/(dx) [(1 + x) ln (1 + x)-(1 + x)] = ln (1 + x)`
	`text(Using)\ d/(dx) uv=uv^{prime}+vu^{prime}`

`text(LHS)`	`= (1+x) xx 1/(1 + x) + ln(1+x)xx1 +-1`
	`= 1+ ln(1+x)-1`
	`= ln(1+x)`
	`=\ text(RHS … as required)`

f.	`text(Area)`	`= int_0^1 f(x)-g(x)\ dx`
		`= int_0^1 (x-(x^2)/2 + (x^3)/3-ln(x+1))\ dx`
		`= [x^2/2-x^3/6 + (x^4)/12-(1 + x) ln (1+x) + (1+x)]_0^1`
		`text{(using part (e) above)}`
		`= [(1/2-1/6 + 1/12-(2)ln2 + 2)-(ln1 + 1)]`
		`= 5/12-2ln2 + 2-1`
		`= 1 5/12-2 ln 2\ \ text(u²)`

Calculus, 2ADV C3 2010 HSC 9b

Let `y=f(x)` be a function defined for `0 <= x <= 6`, with `f(0)=0`.

The diagram shows the graph of the derivative of `f`, `y = f^{prime}(x)`.

The shaded region `A_1` has area 4 square units. The shaded region `A_2` has area 4 square units.

For which values of `x` is `f(x)` increasing? (1 mark)

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What is the maximum value of `f(x)`? (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Find the value of `f(6)`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Draw a graph of `y =f(x)` for `0 <= x <= 6`. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `f(x)\ text(is increasing when)\ 0 <= x < 2`

b. `text(MAX value of)\ f(x) = 4`

c. `-6`

Show Worked Solution

a.	`f(x)\ text(is increasing when)\ \ f^{prime}(x) > 0`
	`text(From the graph)`
	`f(x)\ text(is increasing when)\ 0 <= x < 2`

b.	`f^{prime}(x) = 0\ \ text(when)\ \ x=2`
	`:.\ text(MAX at)\ \ x = 2`
	`int_0^2 f^{prime}(x)\ dx = 4\ \ \ (text(given since)\ A_1 = 4 text{)}`
	`text(We also know)`

`int_0^2\ f^{prime}(x)\ dx`	`= [f(x)]_0^2`
	`= f(2)-f(0)`
	`= f(2)\ \ \ \ text{(since}\ f(0) = 0 text{)}`

`=> f(2) = 4`

♦♦♦ Parts (b) and (c) proved particularly difficult for students with mean marks of 12% and 11% respectively.

`:.\ text(MAX value of)\ \ f(x) = 4`

c.	`int_0^4 f^{prime}(x)`	`= A_1-A_2`
		`=0`

`text(We also know)`

`int_0^4 f^{prime}(x)\ dx`	`= int_2^4 f^{prime}(x)\ dx + int_0^2 f^{prime}(x)\ dx`
	`=[f(x)]_2^4 + 4`
	`= f(4)-f(2) + 4\ \ \ (text(note)\ f(2)=4)`
	`=f(4)`

`=> f(4) = 0`

`text(Gradient)=-3\ text(from)\ \ x = 4\ \ text(to)\ \ x = 6`

`:.\ f(6)`	`=-3 (6- 4)`
	`=-6`

♦♦ Mean mark (d) 28%
EXAM TIP: Clearly identify THE EXTREMES when given a defined domain. In this case, the origin is obvious graphically, and the other extreme at `x=6`, is CLEARLY LABELLED!

Calculus, 2ADV C4 2011 HSC 6c

The diagram shows the graph `y = 2 cos x` .

State the coordinates of `P`. (1 mark)

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Evaluate the integral `int_0^(pi/2) 2 cos x\ dx`. (2 marks)

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Indicate which area in the diagram, `A`, `B`, `C` or `D`, is represented by the integral

`int_((3pi)/2)^(2pi) 2 cos x\ dx`. (1 mark)

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Using parts (ii) and (iii), or otherwise, find the area of the region bounded by the curve `y = 2 cos x` and the `x`-axis, between `x = 0` and `x = 2pi`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Using the parts above, write down the value of

`int_(pi/2)^(2pi) 2 cos x\ dx`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `P(0,2)`

b. `2`

c. `C`

d. `8\ text(u²)`

e. `-2`

Show Worked Solution

a. `y = 2 cos x`

`text(At)\ x = 0`

`y = 2 cos 0 = 2`

`:.\ P(0,2)`

b. `int_0^(pi/2) 2 cos x\ dx`

`= [2 sin x]_0^(pi/2)`

`= [2 sin (pi/2)\ – 2 sin 0]`

`= 2\ – 0`

`= 2`

c. `C`

d.	`text(S)text(ince Area)\ A`	`=\ text(Area)\ C,\ \ text(and)`
	`text(Area)\ B`	`= 2 xx text(Area)\ A`

`:.\ text(Total Area)`	`= 2 + (2xx2) + 2`
	`= 8\ text(u²)`

MARKER’S COMMENT: “Using the parts above” in part (e) was ignored by many students. Important to know when finding an area and evaluating an integral may differ.

e. `int_(pi/2)^(2pi) 2 cos x\ dx`

`= text(Area)\ C-text(Area)\ B`

`= 2-(2 xx 2)`

`=-2`

Calculus, 2ADV C4 2012 HSC 13b

The diagram shows the parabolas `y = 5x-x^2` and `y = x^2-3x`. The parabolas intersect at

the origin `O` and the point `A`. The region between the two parabolas is shaded.

Find the `x`-coordinate of the point `A` (1 mark)

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Find the area of the shaded region. (3 marks)

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a. `4`

b. `64/3 \ text(u²)`

Show Worked Solution

a. `y = x^2-3x\ \ …\ (1)`

`y = 5x-x^2\ \ …\ (2)`

`text(Solve:)\ \ (1)=(2)`

`x^2-3x`	`= 5x-x^2`
`2x^2-8x`	`= 0`
`2x(x-4)`	`=0`
`x`	`= 0 \ text(or) \ 4`

MARKER’S COMMENT: Less errors were made in part (b) by students who simplified the definite integral before integrating, as shown in the worked solution.

`:. x text(-coordinate of) \ A \ text(is) \ 4`

b. `text(Area)`	`= int_0^4 (5x-x^2) dx-int_0^4 (x^2-3x) dx`
	`= int_0^4 (5x-x^2-x^2 + 3x) dx`
	`= int_0^4 (8x-2x^2) dx`
	`= [4x^2-2/3x^3]_0^4`
	`= [(4 xx 4^2)-(2/3 xx 4^3)]`
	`= [ 64-128/3 ]`
	`= 64/3 \ text(u²)`

Calculus, 2ADV C4 2012 HSC 10 MC

The graph of `y = f(x)` has been drawn to scale for `0 <= x <= 8`.

Which of the following integrals has the greatest value?

`int_0^1 f(x) \ dx`
`int_0^2 f(x) \ dx`
`int_0^7 f(x) \ dx`
`int_0^8 f(x) \ dx`

Show Answers Only

`B`

Show Worked Solution

`text(S)text(ince the integrals measure the net area under)`

`text(the graph and above the)\ x text(-axis)\ text{(i.e. below the}`

`x text{-axis is a negative value.)}`

`=> B`

Calculus, 2ADV C4 2013 HSC 14d

The diagram shows the graph `f(x)`.

What is the value of `a`, where `a > 0`, so that `int_-a^a f(x)\ dx = 0`? (1 mark)

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`a=4.5`

Show Worked Solution

`text(If)\ int_-a^a f(x)\ dx =0`

♦♦♦ A devilish 1-mark question mid-paper that had a mean mark of just 12%.
MARKER’S COMMENT: The fact that this question was worth only 1 mark means that it is not necessary for students to show any detailed working.

`text(We know the area below the curve and)`

`text(above the)\ x text(-axis = area above the)`

`text(curve and below the)\ x text(-axis.)`

`text(By inspection, we can see)`

`int_-3^-1 f(x)\ dx=0\ \ text(and)\ \ int_2^3 f(x)\ dx= 0`

`text(We need)\ int_3^a f(x)\ dx + int_-a^-3 f(x)\ dx`	`=-3`
`text(because)\ \ int_-1^2 f(x)\ dx=3`

`=>\ text(S)text(ince areas have height = 1, each)`

`text(must be 1.5 units wide.)`

`:. a = 4.5`

Calculus, 2ADV C4 2013 HSC 13b

The diagram shows the graphs of the functions `f(x) = 4x^3-4x^2 +3x` and `g(x) = 2x`. The graphs meet at `O` and at `T`.

Find the `x`-coordinate of `T`. (1 mark)

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Find the area of the shaded region between the graphs of the functions `f(x)` and `g(x)`. (3 marks)

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a. `T\ text(is)\ (1/2, 1)`

b. `text(Shaded area between the curves is)\ 1/48\ text(u²)`

Show Worked Solution

a. `T\ text(occurs when)\ \ f(x) = g(x)`

`4x^3-4x^2 + 3x`	`=2x`
`4x^3-4x^2 + x`	`=0`
`x(4x^2-4x + 1)`	`=0`
`x(2x-1)^2`	`=0`

`2x-1=0 \ \ => \ x=1/2`

`text(Substitute)\ \ x=1/2\ \ text(into)\ g(x)`

`g(1/2) = 2 xx 1/2 = 1`

`:.\ T (1/2, 1)`

b.	`text(Shaded Area)`	`= int_0^(1/2) (f(x)-g(x)) \ dx`
		`= int_0^(1/2) (4x^3-4x^2 + x) \ dx`
		`= [x^4-4/3x^3 + 1/2x^2]_0^(1/2)`
		`= [((1/2)^4-4/3(1/2)^3 + 1/2(1/2)^2)-0]`
		`= 1/16-1/6 + 1/8`
		`= 1/48\ text(u²)`

`:.\ text(Shaded area between the curves is)\ 1/48\ text(u²)`

Calculus, 2ADV C4 2010 HSC 5c

The diagram shows the curve `y=1/x`, for `x>0`.

The area under the curve between `x=a` and `x=1` is `A_1`. The area under the curve between `x=1` and `x=b` is `A_2`.

The areas `A_1` and `A_2` are each equal to `1` square unit.

Find the values of `a` and `b`. (3 marks)

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Show Answer Only

`a=1/e`

`b=e`

Show Worked Solutions

IMPORTANT: Note when `log_e a=1`, the definition of a log means that `e^1=a`. Many students failed to earn an easy 3rd mark by recognising this.

`int_a^1 1/x \ dx`	`=1`
`[ln x]_a^1`	`=1`
`ln1-lna`	`=1`
`lna`	`=-1`
`:.\ a`	`=e^-1=1/e`

`int_1^b 1/x dx`	`=1`
`[lnx]_1^b`	`=1`
`lnb-ln1`	`=1`
`ln b`	`=1`
`:.b`	`=e`

Calculus, 2ADV C4 2010 HSC 4b

The curves `y=e^(2x)` and `y=e^-x` intersect at the point `(0,1)` as shown in the diagram.

Find the exact area enclosed by the curves and the line `x=2`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

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`1/2e^4+e^-2-3/2\ \ text(u²)`

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MARKER’S COMMENT: The best responses used only a single integral before any substitution as shown in Worked Solutions.

`text(Area)`	`=int_0^2e^(2x)\ \ dx-int_0^2 e^-x\ \ dx`
	`=int_0^2(e^(2x)-e^-x)dx`
	`=[1/2e^(2x)+e^-x]_0^2`
	`=[(1/2e^4+e^-2)-(1/2e^0+e^0)]`
	`=1/2e^4+e^-2-3/2\ \ \ text(u²)`