Question

The value of the integral ∫^1_0(x+1)^2 e^x dx is:

a) (2e-1)

b) (1-2e)

c) (1+2e)

d) None

Solution

∫^1_0 (x+1)^2 e^x dx = ∫^1_0(x^2e^x+e^x+2xe^x)dx

[^1_0 x^2e^x-2∫xe^xdx+e^x+2∫ae^xdx]

= [ex(x^2+1)]_0^1

= e[1+1]-(1)

= (2e-1) Ans 9