Question

If f(x)=integral ^sinx, o on the bottom, root1+t^2 dt and g(y)=integral^y, 3 on the bottom f(x) dx find g''(pi/6) my answer is root15/4 is this correct?

Answer 1

\[
f(x) = \int_0^{\sin x}\sqrt{1+t^2}~dt
\]And\[
g(x) = \int_3^y f(x)~dx
\]Okay, how did you get your answer?

Answer 2

\[
f(x) = \int_0^{\sin x}\sqrt{1+t^2}~dt
\]And\[
g(x) = \int_3^y f(x)~dx
\]Okay, how did you get your answer?

Answer 3

i did root1+(sin pi/6)^2 * cos(pi/6)=
root1+1/4 + root3/2=
root5/4 * root3/2=
root15/4

Answer 4

The issue is, don't we need \(y=h(x)\)?

Answer 5

that's why I'm a bit confused

Answer 6

Oh wait, let me think.

Answer 7

\[
g(y) = \int_3^y f(x)~dx
\]So then:\[
g'(y) = f(y)
\]And:\[
g''(y) = f'(y)
\]Then \[
f(y) =\int_0^{\sin y} \sqrt{1+t^2}~dt
\]So\[
f'(y) = (\cos y)\sqrt{1+\sin^2 y}
\]

Answer 8

I think you are right.

Answer 9

thank you!!