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The question in my textbook asks:

If $f$ is a continuous function such that $$\int\limits_0^x{f(t)dt}=xe^{2x}+\int\limits_0^x{e^{-t}f(t)dt}$$ for all $x$, find an explicit formula for $f(x)$.

My working goes as follows:

I decided to analyse the equation as an integration by parts $\left(\int udv=uv-\int vdu\right)$, so

$uv]_0^x=xe^{2x}\\ \therefore \text{a possible substitution is}\\ \quad u=t,\qquad v=e^{2t}\\ \quad du=dt,\quad dv=2e^{2t}dt$

and

$\int\limits_0^xvdu=\int\limits_0^x{e^{-t}f(t)dt}\\ \therefore e^{-t}f(t)=e^{2t}\\ \quad f(t)=e^{3t}$

This looks sound until I try equating $\int\limits_0^xudv=\int\limits_0^x f(t)dt$, whereupon I get $f(t)=2te^{2t}$.

I think I don't quite understand what an explicit formula is.

ahorn
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3 Answers3

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Hint : All you have to do is differentiate! This is actually pretty much the solution... I let you figure out the rest ; feel free to comment and discuss.

Remark : You do understand what an explicit formula is, you just didn't manage to find it yet. $f(t) = 2t e^{2t}$ is an example of an explicit formula for a function of a real variable.

Hope that helps,

Patrick Da Silva
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Hint: As you have proper condition for the function so you could use Fundamental theorem of calculus for both sides.

Mikasa
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Notice that if you differentiate the equation:

$$f(x) = e^{2x} + 2xe^{2x} + e^{-x}*f(x)$$

$$f(x)(1 - e^{-x}) = e^{2x} + 2xe^{2x}$$

$$f(x) = \frac{e^{2x} + 2xe^{2x}}{1 - e^{-x}}$$

Varun Iyer
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  • I thought about differentiating, but I wasn't sure whether I should differentiate by x or t. I can see now that $\frac{d}{dx}\int\limits_0^xf(t)dt=f(x)-f(0)$ – ahorn Jul 20 '14 at 10:47
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    @ahorn: you can't differentiate with respect to $t$ since the scope of $t$ is restricted to inside the integral. In other words, $t$ is a dummy variable; to see this, we could simply change the $t$ in the integral to $s$ and not change anything outside of the integral. That is, $$\frac{\mathrm{d}}{\mathrm{d}t}\int_0^xf(t),\mathrm{d}t$$ doesn't make sense. – robjohn Jul 20 '14 at 20:58
  • @ahorn : The formula you wrote in your comment is not correct, the correct one is $\frac{d}{dx} \int_0^x f(t) , dt = f(x)$. Think about this. :) – Patrick Da Silva Jul 20 '14 at 22:53
  • @PatrickDaSilva: Can you please explicate your statement? I thought that $\int\limits_0^x f(t)dt = F(x) - F(0) \text{ where } F'(y)=f(y)$ – ahorn Jul 22 '14 at 13:55
  • @ahorn : Yes, that is true! But what is the derivative of the constant $F(0)$? :) It's definitely not $f(0)$! – Patrick Da Silva Jul 23 '14 at 14:55
  • I see. $F(0)$ is always constant. I also just looked at the FTC part 1. Thank you – ahorn Jul 25 '14 at 13:10