1. ## The Substitution Rule

I'm not sure how to finish this problem:

Evaluate the definate integral from 0 to 1 x^2(1 + 2x^3)^5 dx

1. u = 1 + 2x^3

2. du/dx = 6x^2

3. du = 6x^2 dx , du/6 = x^2 dx

x = 0, u = 1 + 2(0)^3 = 1
x = 1, u = 1 + 2(1)^3 = 3

I'm not sure what to do next. What I'm doing is not working. The answer is supposed to be 182/9.

2. Originally Posted by zachb
I'm not sure how to finish this problem:

Evaluate the definate integral from 0 to 1 x^2(1 + 2x^3)^5 dx

1. u = 1 + 2x^3

2. du/dx = 6x^2

3. du = 6x^2 dx , du/6 = x^2 dx

x = 0, u = 1 + 2(0)^3 = 1
x = 1, u = 1 + 2(1)^3 = 3

I'm not sure what to do next. What I'm doing is not working. The answer is supposed to be 182/9.
Hey zachb.

You have the idea right, I'm just really confused on the last two lines!

u = 1+2x^3
(1/6)du = x^2

Now plug these into your original equation so we can find the answer to the integral and then plug in the limits.

x^2(1+2x^3)dx => (1/6)udu

You can integrate that easily!

(1/12)u^2+C, but since in the end it's a definite integral, forget the C.

So now back substitute.

(1/12)u^2 => (1/12)(1+2x^3)^2.

Now evaluate the limits.

Jameson

3. I don't understand what you just did. Can you explain why the answer is supposed to be 182/9? If you don't understand the purpose of the last two lines of my first post then I don't think you understand the problem.

4. Originally Posted by zachb
I'm not sure how to finish this problem:

Evaluate the definate integral from 0 to 1 x^2(1 + 2x^3)^5 dx

1. u = 1 + 2x^3

2. du/dx = 6x^2

3. du = 6x^2 dx , du/6 = x^2 dx

x = 0, u = 1 + 2(0)^3 = 1
x = 1, u = 1 + 2(1)^3 = 3

I'm not sure what to do next. What I'm doing is not working. The answer is supposed to be 182/9.
Jameson forgot that it was (1 + 2x^3)^5, so what he did would be incorrect on that basis. here is the solution

do you know how to evaluate the integral between the limits? aparently for the last two lines you tried to change the limits in terms of u, we don't really have to do that (it's very rare when it is benificial). just change the expression to be in terms of x when we're done and use the same limits.

int{x^2(1 + 2x^3)^5}dx between 0 to 1
let u = 1 + 2x^3
=> du = 6x^2 dx
=> (1/6)du = x^2 dx

so our integral becomes:

(1/6)int{u^5}du = (1/6)[(1/6)u^6]
.......................= (1/36)(1 + 2x^3)^6 evaluated between 0 and 1
.......................= (1/36)[(1 + 2(1)^3)^6] - (1/36)[(1 + 0)^6]
.......................= (1/36)(3^6) - (1/36)
.......................= (1/4)(3^4) - 1/36
.......................= 182/9

I didn't explain much, do you understand what i did?

5. I get it now. I was going by an example in my book that really makes a mess out of the part you just showed me. Thanks for the help.

6. Originally Posted by zachb
I don't understand what you just did. Can you explain why the answer is supposed to be 182/9? If you don't understand the purpose of the last two lines of my first post then I don't think you understand the problem.
I misread the original problem. My method still works fine. It just turns into u^5du.

I understand the problem fine. I'm very able to do any Calc I-III problem. I see that you were trying to rewrite the original limits in terms of u, but I think is just extra work because you can back substitute at the end.

7. Originally Posted by Jameson
I misread the original problem. My method still works fine. It just turns into u^5du.

I understand the problem fine. I'm very able to do any Calc I-III problem. I see that you were trying to rewrite the original limits in terms of u, but I think is just extra work because you can back substitute at the end.
we all know you're great at calc I-III, that's why i said your solution is wrong on the basis that you were finding the antiderivative for 1 + 2x^3 instead of (1 + 2x^3)^5. i used your exact method, so i know you would have been right. don't take it the wrong way Jameson...maybe i could have phrased it better and say you "misread" (which i know is what happened) as opposed to you "forgot," my apologies

8. I appreciate it. I'm just stickin' up for the little math I can do! Kidding. I'm not upset in the slightest. I just didn't want others to think I they couldn't trust my help from that little boo-boo.

Thanks

Oh and no need to aplogize! You were most helpful and I'm going to give you some thanks for your post.