Definite integral evaluation

[Imum Coeli]

QUESTION:
Let F(x) be the definite integral from 0 to x^2 of (t^2+3)/(t+1) dt.
Evaluate the following:
1) F(0)
2) F'(x)
3) F'(1)

NOTES:

Clearly the answer to 1) is zero as any integral from a to a is zero.

For 2) I am having trouble. I am fairly sure that F'(x) is equal to (t^2+3)/(t+1). (Because if F(x)= integral from a to b of f(x) dx then F'(x)=f(x)). So I figure that if I substitute x for t in (t^2+3)/(t+1) then that should be the answer. However, this is not the case. So what am I doing wrong?

For 3) I haven't tried anything because 2) is wrong.

Thanks in advance for any ideas.

 

[HallsofIvy]

QUESTION:
Let F(x) be the definite integral from 0 to x^2 of (t^2+3)/(t+1) dt.
Evaluate the following:
1) F(0)
2) F'(x)
3) F'(1)

NOTES:

Clearly the answer to 1) is zero as any integral from a to a is zero.

For 2) I am having trouble. I am fairly sure that F'(x) is equal to (t^2+3)/(t+1)

On the contrary, that can't be correct because F'(x) is function of x and so cannot be equal to a function of t.

. (Because if F(x)= integral from a to b of f(x) dx then F'(x)=f(x)). So I figure that if I substitute x for t in (t^2+3)/(t+1) then that should be the answer. However, this is not the case. So what am I doing wrong?

You are not using the fact that the integral is to \(\displaystyle x^2\), not x. Use the chain rule:
\(\displaystyle \frac{d}{dx}\int_0^{x^2} \frac{t^2+ 3}{t+1} dt= \left(\frac{d}{du}\int_0^u \frac{t^2+ 3}{t+ 1}dt\right)\frac{du}{dx}\)
where \(\displaystyle u= x^2\).

For 3) I haven't tried anything because 2) is wrong.

Thanks in advance for any ideas.

 

[Imum Coeli]

Thank you. Using that fact then 3) falls right into place.