2
$\begingroup$

$$\int^x_0f(t)\,dt = xe^{2x}+\int^x_0e^{-t}f(t)\,dt$$ Assume $f$ is continous, solve for $f$.

NB! I'm in my first calculus course so nothing too advanced please.


While searching for a name for these equations I found out about Fredholm Integral Equation of the First Kind, which led me to a more specific type called Volterra integral equation. Not very certain how they can aid me in solving this though.

I'm certain it is of the first kind as there is no mention of $f$ outside of the integral signs, and I think it's homogenous, for let $x=0$

$$\int^0_0f(t)\,dt = \int^0_0e^{-t}f(t)\,dt \color{red}{=0}$$

$\color{red}{*}$Red means uncertainty

$\endgroup$
0
3
$\begingroup$

Hint: Using the Fundamental Theorem of Calculus, differentiate both sides with respect to $x$.

Note that if $g(t)$ is continuous then the derivative of $\int_a^x g(t)\,dt$ is $g(x)$.

You will end up with an equation that is easy to solve for $f(x)$.

Remark: This is indeed an instance of an integral equation, but no knowledge of integral equations is required to solve the problem.

$\endgroup$
3
  • $\begingroup$ Question, do I take the derivative with respect to x or t? If I take with respect to t, my equation would state $f(t) = e^{-t}\cdot f(t)$ which means $f(t) = 0$. But if I take with respect to x, then the integral signs wouldn't differentiate, or do they in anyways? $\endgroup$
    – B. Lee
    Nov 30 '14 at 2:37
  • 1
    $\begingroup$ The expression $\int_a^x g(t)\,dt$ is a function of $x$, not of $t$. You take the derivative with respect to $x$. Result is $f(x)=\frac{d}{dx}(xe^{2x})+e^{-x}f(x)$. $\endgroup$ Nov 30 '14 at 2:40
  • $\begingroup$ @XMLParsing Remember The Fundamental Theorem of Calculus $\endgroup$
    – Dylan
    Dec 2 '14 at 0:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.