Fundamental Theorem of Calculus Problem

[Hunny]

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 :(

 

[mfb]

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.

 

[Hunny]

Oh, I didn't even have the theorem correct
Mmm.. I'm not quite following you...

 

[mfb]

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.

 

[Hunny]

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?

 

[Mark44]

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
$g(x) = \int_0^{x^2} f(t)dt = x \ln(x)$

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.

 

[Hunny]

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!

 

[mfb]

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.

 

[Fredrik]

This calculation makes sense too: $$\frac{d}{dx}\int_0^{x^2} f(t)dt =\frac{d(x^2)}{dx}\frac{d}{d(x^2)}\int_0^{x^2} f(t)dt = 2x f(x^2).$$

 

[Mark44]

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.

 

[WWGD]

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.

 

[pasmith]

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.

 

[WWGD]

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.