2 Problems involving integrals

**VectorRun** · December 10th 2010, 10:38 AM

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

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

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

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

My other question is this: evaluate $\lim_{h\to0}} \frac{1}{h}\int_x^{x+h}e^{arctan{t}}dt$ 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 $\lim_{h\to0}}\frac{1}{h}{e^{arctan(x+h)}}$ ? 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?

**HallsofIvy** · December 10th 2010, 11:46 AM

Originally Posted by VectorRun

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

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

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

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 $\lim_{h\to0}} \frac{1}{h}\int_x^{x+h}e^{arctan{t}}dt$ 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 $\lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$ where $f(x)= \int_a^x e^{arctan(t)}dt$ . And, by the Fundamental Theorem of Calculus, $\frac{d}{dx}\int_a^x f(t)dt= f(x)$ .

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 $\lim_{h\to0}}\frac{1}{h}{e^{arctan(x+h)}}$ ? 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.

**Soroban** · December 10th 2010, 11:49 AM

Hello, VectorRun!

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

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

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

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

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

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

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