calculus problem plz help!!?

Favorite Answer

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!...Show more

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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!...Show more

∫(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...Show more

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.................Show more