Test Index

Integral Calculus

© examsnet.com

Question : 65 of 73

Marks: +1, -0

\int x^{2} e^{x} dx

$e^{x}(x^{2}-2 x+2)+c$
$e^{x}(x^{2}-2)+c$
$e^{x}(x^{2}+2 x+2)+c$
$e^{x}(x^{2}-2 x)+c$

Solution:

Concept:

Use integration by parts, choosing

x^2

as the first function and

e^x

as the second function.

Explanation:

Let

I = \int x^2 e^x \, dx

.

Take

u = x^2

,

dv = e^x dx

. Then

du = 2x dx

,

v = e^x

.

Integration by parts gives:

I = x^2 e^x - \int e^x \cdot 2x \, dx = x^2 e^x - 2\int x e^x \, dx

.

Now integrate

\int x e^x \, dx

similarly: let

u = x

,

dv = e^x dx

, so

du = dx

,

v = e^x

.

Thus

\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C_1

.

Substitute back:

I = x^2 e^x - 2(x e^x - e^x) + C = x^2 e^x - 2x e^x + 2e^x + C = e^x(x^2 - 2x + 2) + C

.

Answer:

e^{x}(x^{2}-2 x+2)+c

© examsnet.com

Go to Question:

Prev Question

Next Question