Let $ f(x) $ be the solution of $ \ \int_{0}^{x}e^{(x-t)} f(t) dt=x, \ \ x>0 $. Then $ f(1)= ? $
(a) $-1 $,
(b) $ 0 $,
(c) $ 1 $,
(d) $ 2 $,
Ans. $ \int_{0}^{x}e^{(x-t)}f(t)dt=x $. Differentiating under integral sign , we get
Any help ?
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2Write it as $\int_0^x e^{-t}f(t){\rm d}t = x e^{-x}$. This is easier to differentiate. In general if both the integrand and the limits depend on $x$ then you can use Leibniz rule to find the derivative. – Winther Jun 23 '17 at 00:09
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can you tell me the answer ? – MAS Jun 23 '17 at 00:11
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2Why don't you try to do the work instead! If something is not clear about the derivation then you can ask about that. – Winther Jun 23 '17 at 00:11
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1how to solve , i am not sure . – MAS Jun 23 '17 at 00:12
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To derive the integral function, use the fundamental theorem of calculus – matboy Jun 23 '17 at 00:14
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ok , i got it , f(x)=1-x. – MAS Jun 23 '17 at 00:16
2 Answers
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You are differentiating it wrongly.
After differentiating you should get: $$f(x)+\int_0^x e^{x-t}f(t) \ dt =1$$
Sahiba Arora
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1The integral is a known laplace transform. It's number 33 in http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf – Jun 23 '17 at 00:10
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By differentiation: \begin{align} \frac{d}{dx} \, \int_{0}^{x} e^{x-t} \, f(t) \, dt &= \frac{d}{dx} \, x = 1 \\ \int_{0}^{x} \frac{d}{dx}\left( e^{x-t} \, f(t) \right) \, dt + \frac{d}{dx} \, (x) \, e^{x - x} \, f(x) &= 1 \\ \int_{0}^{x} e^{x-t} \, f(t) \, dt + f(x) &= 1 \\ x + f(x) &= 1 \\ f(x) = 1-x. \end{align}
Alternate method by Laplace transform (standard notation): \begin{align} \mathcal{L}\left\{\int_{0}^{t} e^{t - u} \, f(u) \, du \right\} &= \mathcal{L}\left\{ t \right\} \\ f(s) * \mathcal{L}\left\{ e^{t} \right\} &= \frac{1}{s^2} \\ \frac{f(s)}{s-1} &= \frac{1}{s^2} \\ f(s) &= \frac{1}{s} - \frac{1}{s^2} \\ f(t) &= 1 - t. \end{align}
Now, since $f(t) = 1- t$ the desired result is evident.
Leucippus
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