Question

If F(x) = integral evaluated from 0 to x (sqrt((t^5)+4)dt) then find F'(2). Calc geniuses, help please!

Answer 1

\[F(x)=\int_0^x(\sqrt{t^5}+4)dt\] like that?

Answer 2

or
\[F(x)=\int_0^x\sqrt{t^5+2}dt\] like that

Answer 3

extend the sqrt over the four and you got it!

Answer 4

this one
\[F(x)=\int_0^x\sqrt{t^5+4}dt\]

Answer 5

yes

Answer 6

you are going to be disgusted at how easy this is

Answer 7

by the fundamental theorem of calculus, if
\[F(x)=\int_0^x\sqrt{t^5+4}dt\] then
\[F'(x)=\sqrt{x^5+4}\]

Answer 8

Wow. Well thank you:) and yes all the disgust right now haha

Answer 9

and therefore
\[F'(2)=\sqrt{2^5+4}\] whatever that is

Answer 10

yw