How Do You Solve the Integral of xe^(-x^2) from 0 to 1 Using Substitution?

[hamadee]

hii everyone
I got your forum from the Internet, and noted that you are cooperating with one another
I hope to help me in question
I would be grateful to you

this is my question

∫0^1 xe^(-x)^2 dx


please help me guys

 

[Mark44]

hii everyone
I got your forum from the Internet, and noted that you are cooperating with one another
I hope to help me in question
I would be grateful to you

this is my question

∫0^1 xe^(-x)^2 dx


please help me guys

Is this your integral?
\int_0^1 xe^{-x^2}dx

What have you tried? What integration techniques do you know?

 

[hamadee]

Is this your integral?
\int_0^1 xe^{-x^2}dx

What have you tried? What integration techniques do you know?

yes this is the integral :)

I do not know anything about this question
i tried a lot of thing
but what i know that i should use u and du

:)

 

[Je m'appelle]

yes this is the integral :)

I do not know anything about this question
i tried a lot of thing
but what i know that i should use u and du

:)

Sounds like a substitution problem then.

Try using u=-x^2, rewrite the equation in terms of u and work it out.

 

[hamadee]

Sounds like a substitution problem then.

Try using u=-x^2, rewrite the equation in terms of u and work it out.

∫0^1 xe^(-x)^2 dx

u=-x^2
du=-2x

-1/2 ∫0 to -1 u du

= -1/2 . u^2/2

= -1/2 . (-x^2)^2/2 ]0 to -1

tell me if this true or not

 

[Mark44]

No, this is incorrect. udu -x^2 * (-2xdx). That's not the same as your integrand.

 

[hamadee]

No, this is incorrect. udu -x^2 * (-2xdx). That's not the same as your integrand.

ohhh
okay can you Explain it to me ?!

 

[Mark44]

Your original integrand is xe^-x2dx.

The substitution you are doing is u = -x², du = -2xdx. Use this substitution in your integrand so that it is in terms of u and du, not x and dx.