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

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?

• 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. – John Martin May 16 '16 at 15:59

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.
Write $$F(x) = x \int_0^x f(t) \, dt$$ and use the product rule: $$F'(x) = x f(x) + \int_0^x f(t) \, dt.$$
• What if $f(t)$ is not integrable with respect to $t$? – Puzzled417 May 16 '16 at 16:04
• Then the function $F$ is not defined and hence not differentiable. – Umberto P. May 16 '16 at 16:06
• So this is under the assumption $f(t)$ is integrable with respect to $t$. – Puzzled417 May 16 '16 at 16:17