Question

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

Answer 1

`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`

Answer 2

`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`