Sunday, October 6, 2019

calculus - Proving :$ frac{textrm{d}}{textrm{d}x}int^{g(x)}_{h(x)}f(t)textrm{d}t =f(g(x))g'(x)-f(h(x))h'(x). $



How to prove that :



$$ \frac{\textrm{d}}{\textrm{d}x}\int^{g(x)}_{h(x)}f(t)\textrm{d}t =f(g(x))g'(x)-f(h(x))h'(x). $$


Answer




Let us first assume that $f$ has a primitive, which we shall refer to as $F$. By the fundamental theorem of calculus, we have:



$$\int_{h(x)}^{g(x)}{f(t)\:dt}=F(g(x))-F(h(x))$$



By the chain rule, we have:



$$\frac{d}{dx}\left(f\circ g\right)=f'(g(x))g'(x)$$



As we know that $\frac{d}{dx}F(x)=f(x)$, we have:




$$\frac{d}{dx}\left(F(g(x))-F(h(x))\right)=F'(g(x))g'(x)-F'(h(x))h'(x)\\=f(g(x))g'(x)-f(h(x))h'(x)$$



Which means that:



$$\frac{d}{dt}\int_{h(x)}^{g(x)}f(t)\:dt=f'(g(x))g'(x)-f(h(x))h'(x)$$



Q.E.D.


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