From differentiating xe^x, how do you deduce int_0^2xe^x from that?

2 Answers
Feb 21, 2018

e^2+1 or 8.389

Explanation:

You can't find the integral of a function with its derivative.

We can do this by, differentiating xe^x. Say the derivative is f(x). We can only find the integral of xe^x, using f(x), by saying:

int(intf(x)dx)dx

But that just reduces to:

intxe^xdx.

Let's solve for that, then.

Applying integration by parts, that is:

intuvdx, where u and v are functions, is:

uintvdx-intu'(intvdx)dx

Here, u=x and v=e^x. Inputting:

x inte^xdx-intx'(inte^xdx)dx

x*e^x-inte^xdx

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

Now we solve the definite integral.

Since we have int_0^2xe^xdx, we input:

(e^2*2-e^2+C)-(e^0*0-e^0+C)

(e^2(2-1)+C)-(0-1+C)

e^2+C+1-C

e^2+1, which is equal to:

8.389

Feb 21, 2018

int_0^2 \ xe^x \ dx = e^2+1

Explanation:

Let us start, as indicated, by differentiating the function xe^x using the product rule:

d/dx xe^x = x(d/dxe^x) +(d/dx x)e^x
\ \ \ \ \ \ \ \ \ \ \ = xe^x +e^x

And so we we can write:

xe^x = d/dx(xe^x)-e^x

If we now integrate wrt x we can deduce:

int \ xe^x \ dx = int \ d/dx(xe^x) \ dx - int \ e^x \ dx
\ \ \ \ \ \ \ \ \ \ \ \ \ = xe^x -e^x + C

So we can readily evaluate the definite integral:

int_0^2 \ xe^x \ dx = [xe^x -e^x]_0^2
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (2e^2-e^2) - (0-e^0)
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2e^2-e^2+1
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = e^2+1