Question

can someone walk me through the steps of this, I'm just not quite getting it
Evaluate the integral
\int x^2 (x^3 - 4)^{42} dx ,
by making the substitution u = x^3 -4.

Answer 1

what happens if you differentiate your substitution?

Answer 2

\[\int\limits_{}^{}x^2(x^3-4)^{42} dx\]

Answer 3

Is my question clear to you?

Answer 4

that ends up being 3x^2 I know how to find that I just don't know what to do with all of it

Answer 5

Well, if you look at it closely:

du/dx=3x^2 right?

That almost looks like the x^2 term before the bracket in your integral doesn't it? I am sure with some simple algebraic manipulation you can make it look just like it and then substitute it back in.

Answer 6

I'm lost

Answer 7

The method is called substitution because you want exactly do that, perform an u-substitution, but you cannot just substitute an u in your integral and then happily integrate. You need to differentiate the u term with respect to x too, this will give you:

du/dx=3x^2
Now you want to solve this equation for dx and plug it back into your original equation. You see that at the end of your integral is a dx, and maybe you also understand now why this dx is such important and can't be left away.

Answer 8

Ok so I end up with (1/43)u^43+C
which is the same as the x^2 part right?

Answer 9

or is it the equivalent of 3x^2?

Answer 10

where did the 1/3 go?

Answer 11

maybe that's my problem do I have to put the antiderivative of 3x^2 back in there somewhere?

Answer 12

Yes exactly just what I mentioned above:
du/dx=3x^2

that is the antiderivative of your substitution, you want to substitute that back in too, a substitution ALWAYS requires two substitutions, one substitution attempt, the u-substitution and then right after that differential of u with respect to x.

That's why you need to solve the differential (or the quotient) du/dx=3x^2 for dx, by using regular algebra, and then substitute the dx back into the place where it says dx on your integral (at the very end of it)