# Definite Integrals

1. Feb 12, 2009

### DollarBill

1. The problem statement, all variables and given/known data
I'm not sure if I'm doing this right or not:

If F(x)=$$\int_0^x{\sqrt{t^3+1}dt}$$, then find F'(2)

3. The attempt at a solution
$$\int_0^x$$1/2(t3+1)-1/2*3t2

1/2(x3+1)-1/2*3x2

1/2(23+1)-1/2*3(2)2

2. Feb 12, 2009

### Knissp

Think back to the (first) fundamental theorem of calculus.

By the way, your integration is wrong. What leads you to assume that

$$\int_0^x{\sqrt{t^3+1}dt}$$

is equal to
$$\int_0^x{1/2(t3+1)-1/2*3t2}$$ ??

It looks like you were thinking of a u-substitution, but that doesn't work.
Letting $$u = t^3+1$$, $$du=3t^2 dt$$, so $$dt = \frac{du}{3t^2}$$
You would then have to substitute this dt into your integral for a valid u-sub, but you notice that you would still have an expression in terms of t, so you cannot integrate with respect to u.

3. Feb 12, 2009

### DollarBill

I was thinking of integration by reverse chain rule

4. Feb 12, 2009

### Dick

Reverse chain rule is substitution. It doesn't work. As Knissp said, you can't integrate that function. You don't want the integral anyway, you want the derivative of the integral. Fundamental theorem of calculus.

5. Feb 12, 2009

### DollarBill

Still not sure if I'm doing it right

$$\int_0^x{\sqrt{t^3+1}dt}$$

$${\sqrt{x^3+1}}$$

$${\sqrt{2^3+1}}$$-$${\sqrt{0^3+1}}$$

3-1=2

6. Feb 12, 2009

### w3390

that is correct

7. Feb 12, 2009

### Dick

Uh, not quite. Suppose F'(t)=sqrt(t^3+1), i.e. the antiderivative is F(t). Then the integral from 0 to x of sqrt(t^3-1)=F(x)-F(0), right? Find d/dx of that. The derivative of F(0) is zero, correct? It's a constant.

8. Feb 12, 2009

### Dick

It is not.

9. Feb 13, 2009

### Staff: Mentor

I concur with Dick.

Also, it would be helpful if you identified the things you're working with, rather than write a bunch of disconnected expressions.

$$F(x) = \int_0^x{\sqrt{t^3+1}dt}$$
$$So, F'(x) = \sqrt{x^3+1}$$
So far, so good. Now, what is F'(2)?

10. Feb 13, 2009

### DollarBill

Ok, thanks got the answer (3). Yea, I probably should write out the expressions more often

11. Feb 13, 2009