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Find $F'(x)$ if $F(x) = \displaystyle \int_{0}^x xf(t)dt$.

I tried performing integration by parts to get $\displaystyle F(x) = xtf(t)-\int f'(t)f(x)xt dt$, but I am not sure how that helps. Also, are we assuming $x(f(t))$ to be integrable because otherwise the integral doesn't exist?

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    $\begingroup$ You can write $F(x) = x\int_0^xf(t)dt$ and then just use the product rule. Presumably $x$ is not a function of $t$ and thus you can pull it out of the integral. $\endgroup$ – John Martin May 16 '16 at 15:59
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Note that $x$ is a constant with respect to the variable of integration, so $$\int_{0}^{x}xf(t)\text{ d}t = x\int_{0}^{x}f(t)\text{ d}t\text{.}$$ Notice that $F$ is a product of two functions of $x$: $x$, and $\int_{0}^{x}f(t)\text{ d}t$. So, we need to use the product rule. $$F^{\prime}(x) =(x)^{\prime} \int_{0}^{x}f(t)\text{ d}t+x\left[\int_{0}^{x}f(t)\text{ d}t \right]^{\prime} = \int_{0}^{x}f(t)\text{ d}t+xf(x)\text{.}$$ Note that I used the Fundamental Theorem of Calculus for computing $\int_{0}^{x}f(t)\text{ d}t$. One of the assumptions here, as you can see, is that $f$ needs to be continuous.

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Write $$F(x) = x \int_0^x f(t) \, dt$$ and use the product rule: $$F'(x) = x f(x) + \int_0^x f(t) \, dt.$$

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  • $\begingroup$ What if $f(t)$ is not integrable with respect to $t$? $\endgroup$ – Puzzled417 May 16 '16 at 16:04
  • $\begingroup$ Then the function $F$ is not defined and hence not differentiable. $\endgroup$ – Umberto P. May 16 '16 at 16:06
  • $\begingroup$ So this is under the assumption $f(t)$ is integrable with respect to $t$. $\endgroup$ – Puzzled417 May 16 '16 at 16:17
  • $\begingroup$ Of course it is. $\endgroup$ – Umberto P. May 16 '16 at 16:18

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