I found this formula in my math book : $$\frac{d}{dx} \int _{v(x)}^{u(x)}f(t)dt=u'(x)f(u(x))-v'(x)f(v(x))$$ I have no problem with this formula, but it extension which says if it is possible to have a term $x$ in the fucntion $f$, the formula should be such: $$\frac{d}{dx} \int _{v(x)}^{u(x)}f(t,x)dt=u'(x)f(u(x),x)-v'(x)f(v(x),x)+\int _{v(x)}^{u(x)}\frac{\partial}{\partial x} f(t,x)$$ Where does the third term come from?
Fundamental theorem of calculus and related formulas
1
$\begingroup$
calculus
integration
-
0Do you know how to prove the first formula? If you do, you shouldn't have trouble proving the second one. And if you don't, you should be asking about the first one instead. – 2017-01-29
-
0i'll appreciate if you give the proof – 2017-01-29
-
0This is known as "differentiation under the integral sign." – 2017-01-29