# 2 Problems involving integrals

• Dec 10th 2010, 10:38 AM
VectorRun
2 Problems involving integrals
Suppose $\displaystyle f(0)=0$ and $\displaystyle \int_0^2{f'(2t)e^{f(2t)}}dt=5$. Find the value of f(4).

So I first recognized that $\displaystyle \frac{1}{2}e^{f(2t)}$ should be the antiderivative of the integral by FTC so that $\displaystyle \frac{1}{2}(e^{f(4)}-e^{f(0)})=5$.

Since f(0)=0, the equation becomes: $\displaystyle e^{f(4)}-1=10$

Isolating and then taking the natural log of both sides yields: f(4)=ln11
Is that right?

My other question is this: evaluate $\displaystyle \lim_{h\to0}} \frac{1}{h}\int_x^{x+h}e^{arctan{t}}dt$ It was a bonus question on a quiz and I'm wondering how to do it. The thing sort of looks the definition of the derivative?

The first thing I notice is that the variable in the integrand isn't with respect to x. If I rewrote it through FTC, I'd put $\displaystyle \lim_{h\to0}}\frac{1}{h}{e^{arctan(x+h)}}$? I'm not really sure how to do this question. It sort of looks like it could be in the form of the definition of the derivative?
• Dec 10th 2010, 11:46 AM
HallsofIvy
Quote:

Originally Posted by VectorRun
Suppose $\displaystyle f(0)=0$ and $\displaystyle \int_0^2{f'(2t)e^{f(2t)}}dt=5$. Find the value of f(4).

So I first recognized that $\displaystyle \frac{1}{2}e^{f(2t)}$ should be the antiderivative of the integral by FTC so that $\displaystyle \frac{1}{2}(e^{f(4)}-e^{f(0)})=5$.

Since f(0)=0, the equation becomes: $\displaystyle e^{f(4)}-1=10$

Isolating and then taking the natural log of both sides yields: f(4)=ln11
Is that right?

Yes, that is correct.

Quote:

My other question is this: evaluate $\displaystyle \lim_{h\to0}} \frac{1}{h}\int_x^{x+h}e^{arctan{t}}dt$ It was a bonus question on a quiz and I'm wondering how to do it. The thing sort of looks the definition of the derivative?
Exactly right! That is $\displaystyle \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$ where $\displaystyle f(x)= \int_a^x e^{arctan(t)}dt$. And, by the Fundamental Theorem of Calculus, $\displaystyle \frac{d}{dx}\int_a^x f(t)dt= f(x)$.

Quote:

The first thing I notice is that the variable in the integrand isn't with respect to x.
The variable in the integrand is a "dummy variable"- it could be anything without changing the integral.

Quote:

If I rewrote it through FTC, I'd put $\displaystyle \lim_{h\to0}}\frac{1}{h}{e^{arctan(x+h)}}$? I'm not really sure how to do this question. It sort of looks like it could be in the form of the definition of the derivative?
No, its not f that is being differentiated, it is the integral so the derivative of the integral is the integrand, f.
• Dec 10th 2010, 11:49 AM
Soroban
Hello, VectorRun!

Your first solution is correct . . . ecept for that $\displaystyle \frac{1}{2}$

Quote:

$\displaystyle \displaystyle \text{Suppose }f(0)=0\:\text{ and }\:\int_0^2\!f'(2t)e^{f(2t)}}dt\:=\:5$

$\displaystyle \text{Find the value of }f(4).$

$\displaystyle \text{So I first recognized that }\frac{1}{2}e^{f(2t)}\text{ is the antiderivative of the integral.}$
Um ... not quite.

$\displaystyle \displaystyle \text{We have: }\:\int^2_0 u'\,e^u\,dt \;\;\text{ which equals: }\;e^u\,\bigg]^2_0$

$\displaystyle \text{Hence: }\;e^{f(2t)} \,\bigg]^2_0 \;=\;5 \quad \hdots \;\text{ etc.}$