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Calculus, MET1 2006 HSC 2bi

Find an antiderivative of  `int 1 + e^(7x)`  with respect to `x`.   (1 mark)

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`x + 1/7 e^(7x) + c`

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`int 1 + e^(7x)\ dx = x + 1/7 e^(7x) + c`

Filed Under: Integration (L&E), L&E Integration Tagged With: Band 3, smc-740-10-Exponential (indefinite), smc-748-10-Exponential (indefinite)

Calculus, MET1 2013 ADV 11e

Find  `int e^(4x + 1) dx`   (2 marks)

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`1/4 e^(4x + 1) + c`

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`int e^(4x + 1) dx = 1/4 e^(4x + 1) + c`

Filed Under: Integration (L&E), L&E Integration Tagged With: Band 3, smc-740-10-Exponential (indefinite), smc-748-10-Exponential (indefinite)

Calculus, MET2 2016 VCAA 9 MC

Given that  `(d(xe^(kx)))/(dx) = (kx + 1)e^(kx)`, then  `int xe^(kx) dx` is equal to

  1. `(xe^(kx))/(kx + 1) + c`
  2. `((kx + 1)/k)e^(kx) + c`
  3. `1/k int e^(kx) dx`
  4. `1/k (xe^(kx) - int e^(kx) dx) + c`
  5. `1/k^2 (xe^(kx) - e^(kx)) + c`
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`D`

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♦ Mean mark 41%.
`int (kx + 1)e^(kx) dx` `= xe^(kx) + c_1`
`k int xe^(kx) dx + int e^(kx) dx` `= xe^(kx) + c_1`
`k int xe^(kx) dx` `= xe^(kx) – int e^(kx) dx + c_1`
`:. int xe^(kx) dx` `= 1/k (xe^(kx) – int e^(kx) dx) + c`

 
`=>   D`

Filed Under: Integration (L&E), L&E Integration Tagged With: Band 5, smc-740-10-Exponential (indefinite), smc-740-80-Integration by recognition, smc-748-10-Exponential (indefinite), smc-748-80-Integration by recognition

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