Question

Find the derivatives of the function F(x) = integral from ((-2x+1) to (2x+1) of (t^3)/(sqrt(t^2+9)))dt

Answer 1

\[\frac{d}{dt}(\int\limits_{-2x+1}^{2x+1}\frac{t^3}{\sqrt{t^2+9}} dt)?\]

Answer 2

yep

Answer 3

have fun
break it up into two pieces and then use the chain rule twice. in other words start with
\[\int_{-2x+1}^a\frac{t^3}{\sqrt{t^2+9}}dt+\int_a^{2x+1}\frac{t^3}{\sqrt{t^2+9}}\]
then
\[-\int_a^{-2x+1}\frac{t^3}{\sqrt{t^2+9}}dt+\int_a^{2x+1}\frac{t^3}{\sqrt{t^2+9}}\]

Answer 4

\[\text{ Let } g(t)=\frac{t^3}{\sqrt{t^2+9}} \text{ and let G be antiderivative of g}\]
\[\frac{d}{dt}(G(t)|_{-2x+1}^{2x+1})=\frac{d}{dt}(G(2x+1)-G(-2x+1))\]
\[(2x+1)'g(2x+1)-(-2x+1)'g(-2x+1)\]

Answer 5

\[2g(2x+1)+2g(-2x+1)\]

Answer 6

for the first one replace each t by
\[-2x+1\] and multiply the result by -2
to get
\[\frac{2(-2x+1)^3}{\sqrt{(-2x+1)^2+9}}\]

Answer 7

\[2[\frac{(2x+1)^3}{\sqrt{(2x+1)^2+9}}+\frac{(-2x+1)^3}{\sqrt{(-2x+1)^2+9}}]\]

Answer 8

wait

Answer 9

nvm sorry i thought i entered it wrong

Answer 10

we are using two different methods but it amounts to the same thing

Answer 11

goodnight i'm not simplifying that mess

Answer 12

haha ok i got that far

Answer 13

good night irene

Answer 14

goodnight jack

Answer 15

any tips for simplification?

Answer 16

no simplification. leave it