WILL GIVE A MEDAL AND BECOME A FAN OF

d/dx integral from 0 to 2x of (e^t+2t)dt=
(A) e^2x + 4x
(B) e^2x+ 4x -1
(C) e^2x + 4x^2 -1
(D) 2e^2x + 4x
(E) 2e^2x + 8x

Please show work as well as pick one of the answers!

Question

WILL GIVE A MEDAL AND BECOME A FAN OF

d/dx integral from 0 to 2x of (e^t+2t)dt=
(A) e^2x + 4x
(B) e^2x+ 4x -1
(C) e^2x + 4x^2 -1
(D) 2e^2x + 4x
(E) 2e^2x + 8x

Please show work as well as pick one of the answers!

anonymous · Answer

the derivative of the integral is the integrand, plus chain rule

anonymous · Answer

replace every $t$ in the integrand by $2x$ and then multiply the whole thing by $2$

anonymous · Answer

I'm kind of confused. Can you like show me how to do the work?

anonymous · Answer

where you see a $t$ put $2x$ then multiply by 2 
$$(e^{2x}+2(2x))2$$

anonymous · Answer

and why do you multiply everything by 2?

anonymous · Answer

because the derivative of $2x$ is $2$
you are using the chain rule

anonymous · Answer

oh right lol had a brain fart there

anonymous · Answer

$$F(g(x))=\int_a^{2x}e^t+2tdt$$
$$g(x)=2x, g'(x)=2, [F(g(x))]'=F'(g(x))\times g'(x)$$

anonymous · Answer

thanks!