calculus problem plz help!!?

We'll have to use integration by parts:

S(u dv) = uv - S(v du)

Let:

dv = e^(-x) dx

and

u = x

Thus, v = -e^(-x)

and

du = dx

Substituting into the integration by parts formula:

= uv - S(v du)

= (x)(e^-x) - S(-e^(-x) dx)

= xe^-x + S(e^-x dx)

= xe^-x + -e^-x

= (e^-x)*(x - 1)

Now you can fill in 0 and 1 for x to determine the definite integral, if that's what was required.

hope that helps!

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This problem can be solved using bernoulli formula

∫udv=uv-u'v1....

ur problem

∫ xe^(-x) dx

here u=x and dv=e^(-x)

∫ xe^(-x) dx

integrating the above equation

v=-e^(-x) u=x

v1=-e^(-x) u'=1

v2=-e^(-x) u"=0

the integration ends

the integrated equation is

∫ xe^(-x) dx =e^(-x)-xe^(-x)

substituting the limits

the integrated equation becomes

=[e^(-1)-e^(-1)] - [e^(-0)]

=-1 +c

I hope that i have given u the right answer

Cheers mate!

∫(0 to 1, x e^(-x) dx )

To solve this, you must use integration by parts. You select a u and dv, and use the following formula:

uv - ∫(v du )

With that said,

Let u = x. dv = e^(-x) dx

du = dx. v = (-1)e^(-x)

Using the above formula, our integral becomes

x(-1)e^(-x) - ∫ ( (-1)e^(-x) dx )

Which, if simplified, becomes

-x e^(-x) + ∫ ( e^(-x) dx )

And we can easily integrate this now.

-x e^(-x) + (-1)e^(-x) + C

-x e^(-x) - e^(-x) + C

U CAN SOLVE THIS PROB USING BY PARTS.........

∫ xe^(-x) dx

x∫e^(-x) dx - ∫ (∫e^(-x) dx ){d/dx(x)}dx

-xe^(-x) - ∫ {-e^(-x)}*1*dx

-xe^(-x) + ∫ e^(-x) dx

-xe^(-x) - e^(-x)

e^(-x){-x-1}

then put the value of as the rule of definite integral..............

I have posted all of the steps in handwritten form for ease of understanding at http://fractionaction.blogspot.com/

This might make it easier for you to see the reason for each step.