smc-1055-20-Exponential

Calculus, EXT2 C1 2024 HSC 11a

Find \(\displaystyle \int x e^x\, d x\) (2 marks)

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\(x e^x-e^x+c\)

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\(u=x \quad \ \ u^{\prime}=1\)

\(v^{\prime}=e^x \quad v=e^x\)

	\(\displaystyle\int x e^x \,d x\)	\(=u v^{\prime}-\displaystyle \int v u^{\prime}\, d x\)
		\(=x e^x- \displaystyle \int e^x \cdot 1\, d x\)
		\(=x e^x-e^x+c\)

Which expression is equal to `int x^5 e^{7x} dx`?

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`A`

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`u = x^5`		`v^{′} = e^{7x}`
`u^{′} = 5x^4`		`v = 1/7 e^{7x}`

`int uv^{′}\ dx`	`= uv-int u^{′}v \ dx`
	`= 1/7 x^5 e^{7x}-5/7 int x^4 e^{7x}\ dx`

`=>\ A`

Use integration by parts to find `int x e^(3x) dx`. (2 marks)

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`(x e^(3x))/3-(1)/(9) e^(3x) + C`

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`u`	`= x`	`\ \ \ \ u^{′}`	`= 1`
`v^{′}`	`= e^(3x)`	`v`	`= (1)/(3) e^(3x)`

Find `int x e^(-2x)\ dx.` (3 marks)

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`-1/2 xe^(-2x)-1/4 e^(-2x) + c`

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`text(Integrating by parts:)`

`text(Let)`	`u`	`= x`	`v^{′}`	`= e^(-2x)`
	`u^{′}`	`= 1`	`v`	`= -1/2e^(-2x)`

`int xe^(-2x)\ dx`	`= x · -1/2 e^(-2x) + 1/2int e^(-2x)\ dx`
	`= -1/2 xe^(-2x)-1/4 e^(-2x) + c`

Evaluate `int_0^2 te^-t\ dt.` (3 marks)

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`1-3/e^2`

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`u`	`=t`	`u^{′}`	`=1`
`v`	`=-e^-t\ dt`	`v^{′}`	`=e^-t`

Find `int x e^(2x)\ dx.` (2 marks)

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`(x e^(2x))/2-e^(2x)/4 + c`

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`u`	`=x`	`\ \ \ \ u^{′}`	`=1`
`v^{′}`	`= e^(2x)`	`v`	`= e^(2x)/2`

`:.int xe^(2x)\ dx`	`=x * e^(2x)/2-int e^(2x)/2 * 1\ dx`
	`=(xe^(2x))/2-e^(2x)/4 + c`