How do you find $\int x^{2} e^{1 - x} d x$ $intx^2 e^(1-x)dx$ using integration by parts?

Question

How do you find $\int x^{2} e^{1 - x} d x$ $intx^2 e^(1-x)dx$ using integration by parts?

$\int x^{2} e^{1 - x} d x$ $intx^2 e^(1-x)dx$

The answer is supposedly $- x^{2} e^{1 - x} - 2 x e^{1 - x} - 2 e^{1 - x} + C$ $-x^2 e^(1-x) -2xe^(1-x) -2e^(1-x) + C$ , but I don't understand how it came to be.

Please explain. Thanks in advance!

1 Answer

Impact of this question

1796 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-21T05:34:14

$\int x^{2} e^{1 - x} d x = - x^{2} e^{1 - x} - 2 x e^{1 - x} - 2 e^{1 - x} + C$ $intx^2e^(1-x)dx=-x^2e^(1-x)-2xe^(1-x)-2e^(1-x)+C$

Explanation:

We'll make the following selections:

$u = x^{2}$ $u=x^2$

Take its differential:

$d u = 2 x d x$ $du=2xdx$

$d v = e^{1 - x} d x$ $dv=e^(1-x)dx$

Take its integral:

$v = \int e^{1 - x} d x = - e^{1 - x}$ $v=inte^(1-x)dx=-e^(1-x)$

Apply the integration by parts formula.

$u v - \int v d u = - x^{2} e^{1 - x} - \int - 2 x e^{1 - x} d x$ $uv-intvdu=-x^2e^(1-x)-int-2xe^(1-x)dx$

The negatives cancel each other out.

$= - x^{2} e^{1 - x} + \int 2 x e^{1 - x} d x$ $=-x^2e^(1-x)+int2xe^(1-x)dx$

For $\int 2 x e^{1 - x} d x$ $int2xe^(1-x)dx$ we're going to have to apply integration by parts again, making the following selections:

$u = 2 x$ $u=2x$

$d u = 2 d x$ $du=2dx$

$d v = e^{1 - x} d x$ $dv=e^(1-x)dx$

$v = - e^{1 - x}$ $v=-e^(1-x)$

$u v - \int v d u = - 2 x e^{1 - x} + 2 \int e^{1 - x} d x$ $uv-intvdu=-2xe^(1-x)+2inte^(1-x)dx$

$\int e^{1 - x} d x = - e^{1 - x}$ $inte^(1-x)dx=-e^(1-x)$ , so we get

$\int 2 x e^{1 - x} d x = - 2 x e^{1 - x} - 2 e^{1 - x}$ $int2xe^(1-x)dx=-2xe^(1-x)-2e^(1-x)$

Thus, the entire integral becomes

$\int x^{2} e^{1 - x} d x = - x^{2} e^{1 - x} - 2 x e^{1 - x} - 2 e^{1 - x} + C$ $intx^2e^(1-x)dx=-x^2e^(1-x)-2xe^(1-x)-2e^(1-x)+C$ , adding in the constant of integration.

Science

Math

Humanities

... and beyond

How do you find $\int x^{2} e^{1 - x} d x$ $intx^2 e^(1-x)dx$ using integration by parts?

$\int x^{2} e^{1 - x} d x$ $intx^2 e^(1-x)dx$

The answer is supposedly $- x^{2} e^{1 - x} - 2 x e^{1 - x} - 2 e^{1 - x} + C$ $-x^2 e^(1-x) -2xe^(1-x) -2e^(1-x) + C$ , but I don't understand how it came to be.

Please explain. Thanks in advance!

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find ∫x2e1−xdxintx^2 e^(1-x)dx using integration by parts?

∫x2e1−xdxintx^2 e^(1-x)dx The answer is supposedly −x2e1−x−2xe1−x−2e1−x+C-x^2 e^(1-x) -2xe^(1-x) -2e^(1-x) + C, but I don't understand how it came to be. Please explain. Thanks in advance!

1 Answer

Explanation:

Related questions

Impact of this question

How do you find $\int x^{2} e^{1 - x} d x$ $intx^2 e^(1-x)dx$ using integration by parts?

$\int x^{2} e^{1 - x} d x$ $intx^2 e^(1-x)dx$

The answer is supposedly $- x^{2} e^{1 - x} - 2 x e^{1 - x} - 2 e^{1 - x} + C$ $-x^2 e^(1-x) -2xe^(1-x) -2e^(1-x) + C$ , but I don't understand how it came to be.

Please explain. Thanks in advance!