Answer to Question #147645 in Calculus for liam donohue

Question

If f(x)=∫x0(t3+7t2+4)dt
then
f′′(x)=

Answer 1

By Leibniz's rule, we know that if

$f(x)=\int _ \phi^\psi f(t)dt$

Then $f'(x)=f( \psi(x)).\psi'(x)-f(\phi(x)).\phi'(x)........(1)$

Now given that,$f(x)=\int_0^x (t^3+7t^2+4)......(2)$

Comparing (2) with (1) we get,

$\psi(x)=x ,\phi(x)=0,$

$f(t)=t^3+7t^2+4$

Therefore by Leibniz's rule we have,

$f'(x)=f( \psi(x)).\psi'(x)-f(\phi(x)).\phi'(x)$

So,$f'(x)=(x^3+7x^2+4).(1)-(0).(0) =x^3+7x^2+4$

Therefore, $f''(x)=d/dx(x^3+7x^2+4)=(3x^2+14x)$

So required answer is $f''(x)=3x^2+14x$