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
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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Show Worked Solution
a.
| \(\displaystyle \int_0^6 f(x)\) | \(=\text{net area above the}\ x \text{-axis }\) |
| \(=\text{Area 1}+ \text{Area 3}-\text{Area 2}\) | |
| \(=\left(6+1 \dfrac{1}{2}\right)+3-2\) | |
| \(=8 \dfrac{1}{2}\) |
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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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 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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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
- \(C+D\)
- \(C+D-3\)
- \(C-D\)
- \(D-C\)
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`
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`
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`
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)`
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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Show Answers Only
a. `t > 5\ \ text(seconds)`
b. `7.2\ \ text(seconds)`
c.
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 the area}`
`\text{above the graph from}\ C\ \text(to)\ D, \text(there is no change in)`
`text(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.)`
c.
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?`
- 5
- 9
- 11
- 15
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 2016 HSC 9 MC
What is the value of `int_-3^2 |\ x + 1\ |\ dx?`
- `5/2`
- `11/2`
- `13/2`
- `17/2`
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 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`
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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