Question

Evaluate the integral of (x^2 + 5)e^(-x) dx from 0 to 1.

Answer 1

Explanation:

Let, I=∫(x2+1)e−xdx.

Using the following Rule of Integration by Parts(IBP) :

IBP : ∫uv'dx=uv−∫vu'dx.

We take, u=x2+1,and,v'=e−x.

∴u'=2x,and,v=∫e−xdx=e−x−1=−e−x.

∴I=−(x2+1)e−x−∫(−e−x⋅2x)dx,

⇒I=−(x2+1)e−x+2I1, where, I1=∫xe−xdx.

We once again use IBP for I1; this time, we choose,

u=x,and,v'=e−x∴u'=1,and,v=−e−x.

∴I1=−xe−x−∫(−e−x⋅1)dx,

:I1=−xe−x−e−x.

Utilising I1 in I, we get,

I=−(x2+1)e−x+2{−xe−x−e−x},

⇒I=−(x2+2x+3)e−x+C.

∴∫10(x2+1)e−xdx=−[(x2+2x+3)e−x]10,

=−[(1+2+3)e−1−(0+0+3)e−0].

⇒∫10(x2+1)e−xdx=−(6e−3)=3e(e−2)

Answer 2

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3 f(x) dx −4 where f(x) = 2 if −4 ≤ x ≤ 0 5 − x2 if 0 < 55

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