# Thread: 2 Problems involving integrals

1. ## 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?

2. 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.

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, $\displaystyle \frac{d}{dx}\int_a^x f(t)dt= f(x)$.

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.

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.

3. Hello, VectorRun!

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

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

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

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

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

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