Fundamental Theorem Of Calculus problems help

[raepal]

Fundamental Theorem Of Calculus problems help!

Homework Statement

A))))

Find the derivative of
g(x)=∫[8x to 4x] (u+7)/(u-4) dx

B)))
Use part I of the Fundamental Theorem of Calculus to find the derivative of
h(x) = ∫[sin(x) to -3] (cos(t^3)+t)dt

C)))
F(x) = ∫[ 1 to √3] s^3/(3+5s^4) dx

Homework Equations

The Attempt at a Solution

I tried to do
F(b)b' - F(a)a'
but I am not confident with my answer.

 

[Mark44]

Homework Statement

A))))

Find the derivative of
g(x)=∫[8x to 4x] (u+7)/(u-4) dx

B)))
Use part I of the Fundamental Theorem of Calculus to find the derivative of
h(x) = ∫[sin(x) to -3] (cos(t^3)+t)dt

C)))
F(x) = ∫[ 1 to √3] s^3/(3+5s^4) dx

Homework Equations

The Attempt at a Solution

I tried to do
F(b)b' - F(a)a'
but I am not confident with my answer.

What does part I of the Fundamental Thm. of Calculus say?

 

[Ray Vickson]

Homework Statement

A))))

Find the derivative of
g(x)=∫[8x to 4x] (u+7)/(u-4) dx

B)))
Use part I of the Fundamental Theorem of Calculus to find the derivative of
h(x) = ∫[sin(x) to -3] (cos(t^3)+t)dt

C)))
F(x) = ∫[ 1 to √3] s^3/(3+5s^4) dx

Homework Equations

The Attempt at a Solution

I tried to do
F(b)b' - F(a)a'
but I am not confident with my answer.

Are you sure you mean g(x) = ∫[8x to 4x] (u+7)/(u-4) dx in A)? As written, this means
g(x) = \int_{8x}^{4x} \frac{u+7}{u-4} \, du . (Where you wrote 'dx' I assume you mean 'du'.) The point is: what are the limits, and where they go? If you actually want g(x) = \int_{4x}^{8x} \frac{u+7}{u-4} \, du, you would need to write
∫[4x to 8x] (u+7)/(u-4) du. The answers for g'(x) are different in these two cases. Which one do you really mean?

 

[raepal]

Are you sure you mean g(x) = ∫[8x to 4x] (u+7)/(u-4) dx in A)? As written, this means
g(x) = \int_{8x}^{4x} \frac{u+7}{u-4} \, du . (Where you wrote 'dx' I assume you mean 'du'.) The point is: what are the limits, and where they go? If you actually want g(x) = \int_{4x}^{8x} \frac{u+7}{u-4} \, du, you would need to write
∫[4x to 8x] (u+7)/(u-4) du. The answers for g'(x) are different in these two cases. Which one do you really mean?

It's the second one that you stated.
here is it:

 

[Mark44]

Is this problem B?
You wrote "h(x) = ∫[sin(x) to -3] (cos(t^3)+t)dt"

From your response to Ray, I think this is the function.

$$ h(x) = \int_{-3}^{sin(x)}(cos(t^3) + t)dt$$

Can you work the problem below? This is a little easier, and if you can work it, the one above is only a little harder.
$$ h(x) = \int_{-3}^x(cos(t^3) + t)dt$$

Let me ask again, what does the first part of the Fundamental Thm. of Calculus say?For problem C you wrote "F(x) = ∫[ 1 to √3] s^3/(3+5s^4) dx"

I can only guess at what you meant, which might be this:
$$ F(x) = \int_1^{\sqrt{3}} \frac{s^3}{3 + 5s^4} ds$$

If this is the problem, it's very easy.