1. ## definate integral question

between 0 and t

x^2e^(-x^3) dx

Since t is not determined, I'm guessing the second part of the integral is just going to be variables, but I'm having some difficulty getting there.

For likely the same reasons:

between 0 and 2pi (theta)sin(theta)^2 d(theta) is a bugger. halp?

2. For the first integral, try the substitution $u = -x^3$.

For the second one, try integration by parts, using the reduction formulas for $\int sin^2(\theta) d\theta$ and $\int cos^2(\theta) d\theta$.

3. Originally Posted by JG89
For the first integral, try the substitution $u = -x^3$.

For the second one, try integration by parts, using the reduction formulas for $\int sin^2(\theta) d\theta$ and $\int cos^2(\theta) d\theta$.
We're not into integration by parts, but what about substituting -x^3 for u? Can you describe what I would have to do for that method?

4. If you let $u =- x^3$ then it follows that $\frac{-du}{3} = x^2$.
For the second integral, is it $\int \theta sin^2(\theta) d\theta$ or $\int \theta sin(\theta^2)d\theta$? If it’s the latter, the substitution $u = \theta^2$ should do the trick.

5. So by the power rule for integration I got:

(u^(n+1)) / (n+1) = u^2/2 and therefore the integration of sin is -cos, plug that into u^2/2 and you get (-cosx^2)/2 (x is easier than theta to show)

except I'm not sure where theta in front of the sin in the original problem and since it is between 0 and 2pi, how to get only the area in that range. Sorry if the questions sound dumb, but in this summer class we do a chapter in a day, so there was 5 min spent on this, and I'm afraid it didn't sink in...

6. I found it hard to understand what you are trying to say, but I think I got the gist of it.

We'll use x instead of theta. You're right that an antiderivative of the given integral is $F(x) = \frac{-cos(x^2)}{2}$. To evaluate the area between x = 0 and x = 2pi, then you have to calculate $F(2\pi) - F(0)$.

7. You can use substitution in both cases.

In the first integral, let u=-x^3

so du=-3x^2dx

So the integral reduces to (-e^u)/3 du
with the limits ranging from 0 to -t^3

Hence the answer would be -1/3(e^(t^3) - 1)
=1/3{1-e^(t^3)}