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How may I calculate derivative for the following function?

$$I_n(x)=\int_{a}^{x} (x-t)^{n}f(t)dt$$

Note: I'm given that $f$ is continuous

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  • $\begingroup$ With respect to $x$? $\endgroup$ Jun 19, 2020 at 16:30
  • $\begingroup$ What do you know about the Fundamental theorem of calculus? $\endgroup$ Jun 19, 2020 at 16:31
  • $\begingroup$ $$I_n(x) = \displaystyle\sum_{k=0}^{n}\dbinom{n}{k}(-1)^{n-k}x^k\int_a^xt^{n-k}f(t){\rm d}t.$$ Can you use the Fundamental Theorem of Calculus to conclude now? $\endgroup$ Jun 19, 2020 at 16:32
  • $\begingroup$ @AryamanMaithani Sorry I don't know it and therefore can't use it, I'm looking for a solution using Leibniz integral rule, since I know it will help here but can't implement it :-( $\endgroup$
    – Daniel98
    Jun 19, 2020 at 16:33
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    $\begingroup$ @Daniel98: I suggested FTC because that's the more "basic" rule. If you do know the Leibniz rule, then I'm not sure why you're struggling. I suggest that you edit your question and show us where you got stuck upon applying the rule. (Even if you have no idea, just tell us what you think is the Leibniz rule and why you are not able to apply it.) $\endgroup$ Jun 19, 2020 at 16:37

2 Answers 2

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$$I_n(x)=\int_{a}^{x} (x-t)^{n}f(t){\rm d}t \\ \text{Leibniz rule is-} \\ \dfrac{\rm d}{{\rm d}x}\left(\int_{a(x)}^{b(x)} f(x,t) {\rm d}t\right) = f(x,b(x))\dfrac{\rm d}{{\rm d}x}b(x) - f(x,a(x)\dfrac{\rm d}{{\rm d}x}a(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) {\rm d}t \\ \dfrac{\rm d}{{\rm d}x}I_n(x) = (x-x)^nf(x)\dfrac{\rm d}{{\rm d}x}x - (x-a)^nf(a)\dfrac{\rm d}{{\rm d}x}a + \int_a^xn(x-t)^{n-1}f(t){\rm d}t \\ = nI_{n-1}(x)$$

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  • $\begingroup$ One mistake at the end, it's $I_n$ not $I_{n-1}$ $\endgroup$
    – Daniel98
    Jun 19, 2020 at 17:00
  • $\begingroup$ After taking the partial derivative of $(x-t)^nf(t)$, it became $n(x-t)^{n-1}f(t)$ so there is $n-1$ in the power so it's basically $I_{n-1}$ tell me if you still think there is a mistake $\endgroup$ Jun 19, 2020 at 17:04
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Leibniz's rule tells the following:

$$\dfrac{\rm d}{{\rm d}x}\left(\int_{a(x)}^{b(x)}f(x, t){\rm d}t\right) = f(x, b(x))b'(x) - f(x, a(x))a'(x) + \int_{a(x)}^{b(x)}\left(\dfrac{\partial}{\partial x}f(x, t)\right){\rm d}t.$$


  1. Do you see what $a(x)$, $b(x)$, and $f(x, t)$ will be in this case?
  2. With which of the three terms on the right do you have a problem calculating?
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  • $\begingroup$ How we could know what it x and what is t? I could choose anything... and what if I have 3 variables $\endgroup$
    – Daniel98
    Jun 19, 2020 at 17:01
  • $\begingroup$ Doesn't your question already involve $x$ and $t$? You wish to differentiate $I_n(x)$. Can you try to "match" the appropriate terms? In the case of $I_n(x)$, what appears in the position of $a(x)$? $\endgroup$ Jun 19, 2020 at 17:05
  • $\begingroup$ @Daniel98 the $x$ is the variable wrt which you are taking the derivative $\endgroup$ Jun 19, 2020 at 17:06

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