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Question: If I = \(\int_{- 2}^{0}{\lbrack x^{3} + 3x^{2} + 3x + (x + 1)\cos(x + 1)\rbrack}dx\), then I equals:...

If I = ∫−20[x3+3x2+3x+(x+1)cos⁡(x+1)]dx\int_{- 2}^{0}{\lbrack x^{3} + 3x^{2} + 3x + (x + 1)\cos(x + 1)\rbrack}dx, then I equals:

A

–4

B

–3

C

–2

D

–1

Answer

–2

Explanation

Solution

We can write

I =∫−20[(x+1)3−1+(x+1)cos⁡(x+1)]dx\int_{- 2}^{0}{\lbrack(x + 1)^{3} - 1 + (x + 1)\cos(x + 1)\rbrack}dx

Put x + 1= t, so that

I = ∫−11[t3−1+tcos⁡t]dt\int_{- 1}^{1}{\lbrack t^{3} - 1 + t\cos t\rbrack}dt

= ∫−11(−1)dt\int_{- 1}^{1}{( - 1)}dt = – t]−11\rbrack_{- 1}^{1}= –2

as t3 + t cos t is an odd function