Question

if f is the function given by f(x)= integral from (4 to 2x) of sqrt(t^2-t)dt then f'(2) =?

Answer 1

\[f(x)=\int_4^{2x}\sqrt{t^2-t}dt\] like that?

Answer 2

yea

Answer 3

since the derivative of the integral is the integrand, then by the chain rule
\[f'(x)=\sqrt{(2x)^2-(2x)}\times 2\]

Answer 4

so
\[f'(-2)=\sqrt{(2\times -2)^2-(2\times -2)}\times 2\] whatever that is

Answer 5

im confused we learned we were shown something different.

Answer 6

how would i ido the integral by itself

Answer 7

i think it was substituting x ?

Answer 8

you do not know how to evaluate this integral, but it is not necessary for this problem. you are not asked for the integral, but rather for the derivative of the integral. the derivative if the integral is the integrand i.e. if
\[F(x)=\int_a^xf(t)dt\] then
\[F'(x)=f(x)\]

Answer 9

oh can you show me how to evaluate it though for future reference

Answer 10

in your case you have
\[f(x)=\int_4^{2x}\sqrt{t^2-t}dt\] the derivative is what you get when you replace "t" in the integrand by "2x" , and by the chain rule for derivatives you also have to multiply by the derivative of 2x, which is 2

Answer 11

do you mean how to find the "anti derivative" of
\[\sqrt{t^2-t}\]?

Answer 12

yes

Answer 13

lets say if it was just from (4 to x)

Answer 14

oh heck no. it is a big fat mess forget it

Answer 15

lol

Answer 16

just because you write down a function, doens't mean there is a nice anti derivative of what you write

Answer 17

alright then thanks :)

Answer 18

more often than not you will not be able to find a nice "closed form" for the anti derivative. but you always know that the derivative of the integral is the integrand, so it is always true that if
\[F(x)=\int_a^xf(t)dt\] then
\[F'(x)=f(x)\]