Solve Definite Integrals: Find F'(2)

In summary, the conversation is about finding the derivative of the integral F(x)=\int_0^x{\sqrt{t^3+1}dt}, with the given function F'(x)=\sqrt{x^3+1}. The correct answer for F'(2) is 3, and it is recommended to write out the expressions more clearly.
  • #1
DollarBill
44
0

Homework Statement


I'm not sure if I'm doing this right or not:

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

The Attempt at a Solution


[tex]\int_0^x[/tex]1/2(t3+1)-1/2*3t2

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

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

Answer=2
 
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  • #2
Think back to the (first) fundamental theorem of calculus.

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

[tex]
\int_0^x{\sqrt{t^3+1}dt}
[/tex]

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

It looks like you were thinking of a u-substitution, but that doesn't work.
Letting [tex]u = t^3+1[/tex], [tex]du=3t^2 dt[/tex], so [tex]dt = \frac{du}{3t^2}[/tex]
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
I was thinking of integration by reverse chain rule :confused:
 
  • #4
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
Still not sure if I'm doing it right

[tex]\int_0^x{\sqrt{t^3+1}dt}[/tex]

[tex]{\sqrt{x^3+1}}[/tex]

[tex]{\sqrt{2^3+1}}[/tex]-[tex]{\sqrt{0^3+1}}[/tex]

3-1=2
 
  • #6
that is correct
 
  • #7
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
w3390 said:
that is correct

It is not.
 
  • #9
DollarBill said:
Still not sure if I'm doing it right

[tex]\int_0^x{\sqrt{t^3+1}dt}[/tex]

[tex]{\sqrt{x^3+1}}[/tex]

[tex]{\sqrt{2^3+1}}[/tex]-[tex]{\sqrt{0^3+1}}[/tex]

3-1=2
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.

[tex]F(x) = \int_0^x{\sqrt{t^3+1}dt}[/tex]
[tex]So, F'(x) = \sqrt{x^3+1}[/tex]
So far, so good. Now, what is F'(2)?
 
  • #10
Ok, thanks got the answer (3). Yea, I probably should write out the expressions more often
 
  • #11
Yea, I probably should write out the expressions more often

should read

Yea, I should write out the expressions more often.
 

Related to Solve Definite Integrals: Find F'(2)

1. What is a definite integral?

A definite integral is a mathematical concept used to calculate the area under a curve or the net area between a curve and the x-axis within a specific interval. It is represented by the symbol ∫ and has a lower and upper limit of integration.

2. What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration, while an indefinite integral does not. Additionally, the definite integral gives a numerical value, while the indefinite integral gives a function.

3. How do you solve a definite integral?

To solve a definite integral, you must first find the antiderivative of the function being integrated. Then, you can plug in the upper and lower limits of integration and find the difference between the two resulting values.

4. What is the meaning of F'(2) in the context of a definite integral?

F'(2) represents the derivative of the function F(x) at the point x=2. In the context of a definite integral, it is used to find the slope of the tangent line to the curve at x=2.

5. What are some common techniques for solving definite integrals?

Some common techniques for solving definite integrals include substitution, integration by parts, and trigonometric substitution. Other methods such as partial fractions and using tabular integration tables can also be used for more complex integrals.

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