Standard Integration (Adv-X)

Calculus, 2ADV C4 EO-Bank 12

Find `int 1/((2x)^3)\ dx`. (2 marks)

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`-1/(16x^2) + C`

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`int 1/((2x)^3)\ dx`	`= 1/8 int x^-3\ dx`
	`= 1/8 xx 1/-2 xx x^-2 + C`
	`= – 1/(16x^2) + C`

Calculus, 2ADV C4 2011 HSC 4d v1

Differentiate `y=sqrt(16 -x^2)` with respect to `x`. (2 marks)

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Hence, or otherwise, find `int (8x)/sqrt(16 -x^2)\ dx`. (2 marks)

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a. `- x/sqrt(16-x^2)`

b. `-8 sqrt(16-x^2) + C`

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a. `y= sqrt(16-x^2)= (16-x^2)^(1/2)`

`dy/dx`	`=1/2 xx (16-x^2)^(-1/2) xx d/dx (16-x^2)`
	`= 1/2 xx (16-x^2)^(-1/2) xx -2x`
	`= – x/sqrt(16-x^2)`

b.	`int (8x)/sqrt(16-x^2)\ dx`	`= -8 int (-x)/sqrt(16-x^2)\ dx`
		`= -8 (sqrt(16-x^2)) + C`
		`= -8 sqrt(16-x^2) + C`

Calculus, 2ADV C4 EO-Bank 11

Differeniate \(y=\dfrac{x}{x^2+1}\) (2 marks)

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Hence evaluate \(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\) (2 marks)

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a. \(\dfrac{dy}{dx} = \dfrac{1-x^2}{(x^2+1)^2}\)

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

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a. Using the quotient rule:

\(\dfrac{dy}{dx} = \dfrac{x^2+1-2x^2}{(x^2+1)^2} = \dfrac{1-x^2}{(x^2+1)^2}\)

b. Using part a:

\(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\)	\(=\left[\dfrac{x}{x^2+1}\right]_0^1\)
	\(=\dfrac{1}{2} -0\)
	\(=\dfrac{1}{2}\)

Calculus, 2ADV C4 2010 HSC 2di v1

Find `int sqrt(4x+3) \ dx .` (2 marks)

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` ((4x+3)^(3/2))/6 + C`

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` int sqrt( 4x+3 ) \ dx`	`= 1/(3/2) xx 1/4 xx (4x+3)^(3/2) +C`
	`= ((4x+3)^(3/2))/6 + C`

Calculus, 2ADV C4 EO-Bank 10

Given that `int_0^k ( 2x + 4 )\ dx = 21`, and `k` is a constant, find the value of `k`. (2 marks)

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`k = 3`

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`int_0^k ( 2x + 4 ) \ dx`	`= 21`
`int_0^k ( 2x + 4 ) \ dx`	`= [ x^2 + 4x ]_0^k`
	`= [(k^2 + 4k ) – 0 ]`
	`= k^2 + 4k`

`=> k^2 + 4k`	`=21`
`k^2+4k-21`	`= 0`
`(k-3)(k+7)`	`= 0`
`k`	`=3, -7`
`k`	`=3\ text(as ) k >0`

Calculus, 2ADV C4 2022 HSC 6 MC v1

What is `int(3)/((5x-2)^(2))\ dx` ?

`(-3)/(5x-2)+C`
`(-3)/(5(5x-2))+C`
`(3)/(5) text{ln}(5x-2)+C`
`(3)/(5x-2)+C`

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`B`

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`int 3(5x-2)^(-2)`	`=(3(5x-2)^(-1))/((-1)(5))+C`
	`=(-3)/(5(5x-2))+C`

`=>B`

Calculus, 2ADV C4 2024 HSC 5 MC v1

What is \( {\displaystyle \int x(3 x^2+1)^4 d x} \) ?

\( \dfrac{1}{30}(3x^2+1)^5+C \)
\( \dfrac{1}{5}(3x^2+1)^5+C \)
\( \dfrac{5}{6}(3x^2+1)^5+C \)
\( \dfrac{6}{5}(3x^2+1)^5+C \)

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\( A \)

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\[ \int x(3x^2+1)^4 dx\]	\(=\dfrac{1}{5} \cdot \dfrac{1}{6x}x(3x^2+1)^5+C\)
	\(=\dfrac{1}{30}(3x^2+1)^5+C\)

\( \Rightarrow A \)

NOTE: Integrating by the reverse chain rule only works if some form of the derivative is already present outside of the brackets.

Calculus, 2ADV C4 EO-Bank 4 MC SJ

Let `f^(')(x)=(2)/(sqrt(2x-3))`.

If `f(6)=4`, then

`f(x)=2sqrt(2x-3)`
`f(x)=sqrt(2x-3)-2`
`f(x)=2sqrt(2x-3)-2`
`f(x)=sqrt(2x-3)+2`

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`=>C`

Show Worked Solution

`f^{‘}(x)`	`=2/(sqrt(2x-3))`
`f(x)`	`=2 int(2x-3)^{- 1/2}`
	`=21/22(2x-3)^{1/2}+c`
	`=2sqrt(2x-3)+c`

`text(When)\ \ x=6, \ f(x)=4:`

`4=2sqrt(12-3) + c \ => \ c=-2`

`:. f(x) = 2sqrt(2x-3)-2`

`=>C`