Question

What is `f(x) = int 3x^3-2x+x^2e^(3x-2) dx` if `f(1) = 3`?

Answer 1

`3/4x^4 - x^2 -1/27e^(3x-2)(9x^2 - 6x + 2) + 3.754`
Your teacher is a sadist.

Explanation:

Man, this one is just brutal.

Since its the integral of the sum of 3 terms, you can chop it up into bits, that can be integrated separately. So you'd have:

`int(3x^3)dx - int(2x)dx + int(x^2*e^(3x-2))dx`

The first two integrals are easy...

`= 3/4x^4 - x^2 - int(x^2*e^(3x-2))dx` <- call this eq. 1

leaving only that irritating last integral, which we can attack separately.

You can use integration by parts. But you have to lather, rinse, and repeat (do it twice.)

For integration by parts, remember: if you have u and v, both functions of x, then:

`int(uv')dx = uv - int(u'v)dx`

So, for `int(x^2*e^(3x-2))dx`, we'll let `x^2` be u, and `e^(3x-2)` be v'.

`uv` is then `x^2 * 1/3e^(3x-2)`. u'v is then `2x * 1/3e^(3x-2)`

We can then write:

`int(x^2*e^(3x-2))dx = x^2 * 1/3e^(3x-2) - int(2x * 1/3e^(3x-2))dx`

`= 1/3x^2e^(3x-2) - 2/3int(xe^(3x-2))dx` <- call this eq. 2

...closer. We've got to use integration by parts one more time on the remaining integral.

This time, we let x be u, and `e^(3x-2)` once again be v'. So then, write:

`int(xe^(3x-2))dx = x* 1/3e^(3x-2)-int(1/3e^(3x-2))dx`

...and we can now solve the remaining integral:

`1/3inte^(3x-2)dx = 1/9e^(3x-2) + c`

So we can substitute all this back into eq. 2:

`int(x^2*e^(3x-2))dx = 1/3x^2e^(3x-2) -2/3( x* 1/3e^(3x-2) - 1/9e^(3x-2)) + c`

`= 1/3x^2e^(3x-2) -2/9xe^(3x-2) + 2/27e^(3x-2) + c`

...I cheated & checked this with wolfram. It tells me the preferred way to write this is:

`1/27e^(3x-2)(9x^2 - 6x + 2) + c`

...and we can now substitute THIS all the way back in eq. 1:

`3/4x^4 - x^2 - int(x^2*e^(3x-2))dx =`

`3/4x^4 - x^2 -1/27e^(3x-2)(9x^2 - 6x + 2) + c`

...then we can evaluate this solution at x = 1, solve for c, and knock off for the day.

We're told that f(1) = 3. So

`3/4 - 1 - (1/27 * 2.72(9-6+ 2)) + c = 3`

`-0.754 + c = 3`

c = 3.754