Fundamental Theorem of Calculus Problem

In summary, the FTC states that if a function g(x) is composed of two functions f(x) and g(x), then the derivative of g(x) is given by:
  • #1
Hunny
4
0
1. If g(x) = ∫ f(t) dt = xln x, find f(1)
The ∫ has x^2 on top and 0 on bottom.


2. g'(x) = f(x) <--FTC1

The Attempt at a Solution


g'(x) = f(x) u=x^2
g'(x) = u*lnu * 2x(derivative of inner function)
g'(x) = 2x(x^2)ln(x^2)
f(1) = 2(1)(1^2)ln(1^2)
f(1) = 0, since ln(1) = 0

I keep getting 0, and I'm not sure how 0 is not the answer... The answer key says the solution is 1/2. I really don't know what I'm doing wrong :(
 
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  • #2
Hunny said:
g'(x) = f(x) <--FTC1
That's not what the theorem says. You would need x as upper integration limit to do that.
You can introduce a new variable to fix that, like your u. Plug it in everywhere in the first equation, then proceed as before.
 
  • #3
Oh, I didn't even have the theorem correct :frown:
Mmm.. I'm not quite following you...
 
  • #4
u=x2
You can solve for x and replace all occurences of x in the initial equation. Then it has the right shape to use the theorem because the upper integral limit is the variable itself, not something else.
 
  • #5
I'm going to type out my work, okay? Can you check my work and see if I am doing it correctly?

u = x^2 --> sqrt(u) = x

g'(x) = ∫ u*lnu
with sqrt(u) being the upper limit and 0 being lower

Am I doing it correctly?
 
  • #6
Hunny said:
1. If g(x) = ∫ f(t) dt = xln x, find f(1)
The ∫ has x^2 on top and 0 on bottom.
In a lot nicer form, this is
##g(x) = \int_0^{x^2} f(t)dt = x \ln(x)##
Hunny said:
2. g'(x) = f(x) <--FTC1

The Attempt at a Solution


g'(x) = f(x) u=x^2
g'(x) = u*lnu * 2x(derivative of inner function)
g'(x) = 2x(x^2)ln(x^2)
f(1) = 2(1)(1^2)ln(1^2)
f(1) = 0, since ln(1) = 0

I keep getting 0, and I'm not sure how 0 is not the answer... The answer key says the solution is 1/2. I really don't know what I'm doing wrong :(
Since you're taking the derivative with respect to x, it would help if you kept track of what you're given (the equation I wrote above).

##g'(x) = d/dx (\int_0^{x^2} f(t)dt )= d/dx(x \ln(x))##
To take the derivative in the middle expression, you need to use the chain rule.
The derivative of the expression on the right is straightforward, but you should continue in the pattern above until you get a formula for f.
 
  • #7
I SOLVED IT!
Thank you so much, mfb and Mark44!

I didn't know that I had to find the derivative for xlnx!
Thank you, thank you, thank you!
 
  • #8
That's what I would have done:

$$\int_0^{u} f(t)dt = \sqrt{u} \ln(\sqrt{u}) = \frac{1}{2} \sqrt{u} \ln(u)$$
Using the derivative with respect to u on both sides:
$$f(u) = \frac{d}{du} \left(\frac{1}{2} \sqrt{u} \ln(u)\right)$$
.. and that is quick to calculate and gives f(1)=1/2.
 
  • #9
This calculation makes sense too: $$\frac{d}{dx}\int_0^{x^2} f(t)dt =\frac{d(x^2)}{dx}\frac{d}{d(x^2)}\int_0^{x^2} f(t)dt = 2x f(x^2).$$
 
  • #10
Fredrik said:
This calculation makes sense too: $$\frac{d}{dx}\int_0^{x^2} f(t)dt =\frac{d(x^2)}{dx}\frac{d}{d(x^2)}\int_0^{x^2} f(t)dt = 2x f(x^2).$$
That's more or less what I said in post #6.
 
  • #11
In a generalized way, the FTC deals with composite functions, i.e., (EDIT, from Pasmith's post) ##\frac {d}{dx} \int_{h(x)}^{g(x)} f(t)dt= f(g(x))g'(x)-f(h(x))h'(x) ##, by the chain rule.
 
Last edited:
  • #12
WWGD said:
In a generalized way, the FTC deals with composite functions, i.e., ##\frac {d}{dx} \int_{h(x)}^{g(x)} f(t)dt= f(h(x))h'(x)-f(g(x))g'(x) ##, by the chain rule.

You have a sign error: The lower limit's derivative should be multiplied by -1, not the upper limit's.
 
  • #13
pasmith said:
You have a sign error: The lower limit's derivative should be multiplied by -1, not the upper limit's.

Bah, f(x), -f(x), potato, potatoe :) . Just edited, thanks.
 

FAQ: Fundamental Theorem of Calculus Problem

What is the Fundamental Theorem of Calculus Problem?

The Fundamental Theorem of Calculus is a fundamental concept in calculus which states that the integral of a function can be calculated by finding the antiderivative of that function at the limits of integration.

What is the significance of the Fundamental Theorem of Calculus Problem?

The Fundamental Theorem of Calculus is significant because it provides a powerful tool for evaluating integrals, which are essential in many areas of mathematics and science. It also provides a connection between integration and differentiation, two fundamental concepts in calculus.

What is the difference between the first and second parts of the Fundamental Theorem of Calculus Problem?

The first part of the Fundamental Theorem of Calculus states that the definite integral of a function can be calculated by finding the antiderivative of that function at the limits of integration. The second part states that the derivative of the integral of a function is equal to the original function.

How is the Fundamental Theorem of Calculus Problem used in real-world applications?

The Fundamental Theorem of Calculus is used in various fields such as physics, engineering, economics, and statistics to solve problems involving rates of change and accumulation. It is also used in the development of mathematical models and in the analysis of data.

What are some common mistakes made when solving problems involving the Fundamental Theorem of Calculus?

Some common mistakes include incorrectly identifying the limits of integration, not taking into account the constant of integration, and making errors in finding the antiderivative. It is important to carefully check the work and understand the problem before applying the theorem.

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