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Calculus, SPEC2 2023 VCAA 10 MC

If  \(I_n=\displaystyle {\int_0^1\left((1-x)^n e^x\right) d x}\), where  \(n \in N\), then for  \(n \geq 1, I_n\) equals

  1. \(-1+n I_{n-1}\)
  2. \(n I_{n-1}\)
  3. \(-1-n I_{n-1}\)
  4. \(-n I_{n-1}\)
  5. \((1-x)^n e^x+n I_{n-1}\)
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\(A\)

Show Worked Solution

\(\text{Using integration by parts:}\)

\begin{aligned}
u &=(1-x)^n, & u^{′} &=n(1-x)^{n-1}\\
v^{′} & = e^x, & v & = e^x\  \\
\end{aligned}

\begin{aligned}
\int_0^1 ((1-x)^n e^x)\,dx\ &= uv-\int u^{′} v \ dx\\
& = \Bigr[(1-x)^n\,e^x\Bigr]_0 ^1- \int_0 ^1 n(1-x)^{n-1}\,e^x\ dx  \\
& =-1-n\int_0 ^1 (1-x)^{n-1}\,e^x\ dx  \\
& =-1-nI_{n-1}  \\
\end{aligned}

\(\Rightarrow A\)

Calculus, EXT2 C1 2021 HSC 2 MC

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

  1. `1/7 x^5 e^{7x} - 5/7 int x^4 e^{7x} dx`
  2. `1/7  x^5 e^{7x} - 5/7 int x^5 e^{7x} dx`
  3. `5/7 x^4 e^{7x} - 5/7 int x^4 e^{7x} dx`
  4. `5/7  x^4 e^{7x} - 5/7 int x^5 e^{7x} dx`
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`A`

Show Worked Solution
`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`

Calculus, EXT2 C1 2004 HSC 1a

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`

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

 

`int x e^(3x)  dx` `= uv-int u^{′} v \ dx`
  `= (xe^(3x))/(3)-(1)/(3) int e ^(3x)  dx`
  `= (x e^(3x))/3-(1)/(9)  e^(3x) + C`

Calculus, EXT2 C1 2016 HSC 11b

Find  `int x e^(-2x)\ dx.`  (3 marks)

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

Show Worked Solution

`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`