I have to integrate
(t+1)(t^2+4)
from -3^1/2 to +3^1/2
I know the answer comes out to 10(3)^1/2, but I just can't get it. I tried re-writing it as the integral of t^3+t^2+4t+4 and that just made everything a mess. I also tried to do the integral of (t+1) time the integral of (t^2+4). However, I'm not sure if you can even do that. Can anyone help me with this problem?
Also, this one is giving me trouble (substitution). I'm not sure if I did it right..
Evaluate
10x^1/2
--------------dx
(1+x^3/2)^2
So I made
u=1+x^3/2
du=3/2x^1/2
So then I know that 20/3 will get me what I need.. so..
20/3 integral u^-2
20/3 u^-1/-1
Put it back in the original:
20/3 ((10x^1/2)(1+x^3/2)^-1
--------------------------
-1
Am I on the right track?
Okay I got the first one.
On the second one:
20/3 ((10x^1/2)(1+x^3/2)^-1 (*)
--------------------------
-1
Don't I have to put the 10x^1/2 back in? Like you substitute out something integrate with respect to u then put it back in the original?
And the -1 (I thought) belonged there. Now I'm just confused
The -1 has to be on the bottom so it can get rid of the -1 when it comes down?
and to be more clear this is what I have
20 ((10x^1/2)(1+x^3/2)^-1 (*)
--------------------------
(3) -1
?
So I'm doing my substitution incorrectly.
Up to
20/3 integral u^-2
I integrate that.
20/3 u^-1
So I plug back in 1+x^3/2 into where I see u, then put a one over it and move it to the denominator.
which gives me:
20 (1)
-----------------
3 (1+x^3/2)
So now I'm just confused where you got the negative from?
Yes from the bottom. Thanks.
I have one more question
I have a problem where it wants me to find average value (where you integrate then divide by the length, in this case 1)
h(x) = -|x| on -1, 0 and a few others.
I'm just not sure how to do the integral of that. If I make it 1/2x^2 I won't get the abs value... not sure where to go on this?