Fundamental Theorem of Calculus Problem

  • #1
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 :(
 

Answers and Replies

  • #2
34,976
11,164
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
4
0
Oh, I didn't even have the theorem correct :frown:
Mmm.. I'm not quite following you...
 
  • #4
34,976
11,164
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
4
0
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
34,310
5,949
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
4
0
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
34,976
11,164
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
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
412
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
34,310
5,949
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
WWGD
Science Advisor
Gold Member
2019 Award
5,412
3,488
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
pasmith
Homework Helper
1,879
527
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
WWGD
Science Advisor
Gold Member
2019 Award
5,412
3,488
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.
 

Related Threads on Fundamental Theorem of Calculus Problem

Replies
2
Views
2K
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
784
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
1
Views
978
  • Last Post
Replies
3
Views
803
  • Last Post
Replies
2
Views
604
  • Last Post
Replies
4
Views
1K
Top