Question

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

g(s) = integral of s to 2 of (t − t^9)^3dt

Answer 1

\[\int_s^2(t-t^9)^3dt=-\int_2^s(t-t^9)^3dt\] is a start
then since the derivative of the integral is the integrand, replace all \(t\) by \(s\) and you are dond

Answer 2

*done

Answer 3

\[g(s)=\int\limits_{2}^{s}(t-t^9)^3dt\]

Answer 4

that was the starting equation

Answer 5

so you're saying insert S everywhere t is? then what?

Answer 6

then you are done

Answer 7

what about the 2 in the integral?

Answer 8

the derivative of the integral is the integrand, that is all
the 2 determines the constant, but the derivative of a constant is zero, so it doesn't change anything

Answer 9

ohhhh you're right! i didn't catch that part! thanks alot satellite!

Answer 10

yw

Answer 11

if you don't mind... can you explain to me the difference between integrand and integral?
are they essentially the same thing?

Answer 12

?

Answer 13

\[\huge \int\limits_{s}^{2}\color{red}{(t-t^9)^3}dx\]
The whole expression is an integral.
Red part is the integrand.

Answer 14

oh word! that makes sense... so the derivative of an integral is just the integrand...

Answer 15

Much like how...
\[\huge \sqrt{\color{red}{x^2+6x+9}}\]
the whole expression is a radical, but the red part is called the radicand.

Answer 16

wow that makes a lot of sense actually

Answer 17

Derivative of an integral is the integrand? That makes sense if it were an indefinite integral, or a definite integral of the form...
\[\huge \int\limits_{a}^{x}f(t) dt\]
Then certainly,
\[\huge \frac{d}{dx} \int\limits_{a}^{x}f(t) dt = f(x)\]

Answer 18

so that's what happened with my problem right... change it to s and the derivative of the integral was the integrand of f(s)

Answer 19

Yes. Exactly :)

Answer 20

omg i feel so much smarter!

Answer 21

Except, in your case, it s was the lower limit instead of the upper limit... in that case, since...
\[\huge \int\limits_{s}^{2}f(t) dt=-\int\limits_{2}^{s}f(t) dt\]
then...
\[\huge \frac{d}{ds}\int\limits_{s}^{2}f(t) dt=-f(s)\]

Answer 22

oh okay! makes sense. thanks terenzreignz