# Find the derivative of $F(x) = \int_0^x xf(t) dt$

this was given as an exercise in my first year honours math class. I can't seem to wrap my head around why this is not equal to $xf(t)$. Any help is appreciated! heres the question:

Find the derivative of $F(x) = \int \limits_0^x xf(t) dt$

• Factor out the $x$ from the integral. Use the product rule. Commented Jan 24, 2014 at 15:28
• Or use Leibniz's integral rule. Commented Jan 24, 2014 at 15:29
• @DavidMitra would the answer be xf(x) ? Commented Jan 24, 2014 at 21:34
• No. Let $G(x)=\int_0^x f(t)\, dt$. Then $F(x)=x G(x)$ and, by the product rule, $$F'(x) = [ x] ' \cdot G(x)+G'(x)\cdot x =1\cdot G(x)+ f(x)\cdot x =\int_0^x f(t) \,dt +x f(x).$$ Commented Jan 24, 2014 at 21:36

$x$ is 'constant' with respect to $t$ here - it can take fixed, specific values which have nothing to do with $t$. If it helps, write, say, $$F(3)=\int_{0}^{3}3f(t)dt$$ Now obviously we can factor $3$ out of the integral. Similarly, in general, $$F(x)=x\int_{0}^{x}f(t)dt$$ Now use the product rule and FTC.