Question

if f(x) = the integral from 0 to sinx of radical 1+tsquared dt and g(y) = integral from 3 to y f(x)dx find g''(pi/6)

Answer 1

for f(x), 1+t^2 is all under a radicand

Answer 2

otherwise right!!

Answer 3

\[ f(x) = \int_{0}^{sin(x)} \sqrt{1+t^2}dt\]\[g(y) =\int_{3}^{y} f(x)dx\]

Answer 4

Wth d you do ;_; bah, let me think..

Answer 5

I'm not 100% positive on this, but this is just what I think. So if you take the derivative of g(y) with respect to y, it becomes one of the fundamental theorem of calculus rules.
\[\frac{ d }{ du}\int\limits_{a}^{u}f(t)dt = f(u)du\] The notation on that may not be perfect, but that's the idea. When you differentiate an integral like you have, you exchange the variable you have with the variable expression on the integrand and then multiply by the derivative of the expression on the integrand. In the end, it looks like this:

\[g(y) = \int\limits_{3}^{y}f(x)dx \implies g'(y) = \frac{ d }{ dy }\int\limits_{3}^{y}f(x)dx = f(y)*1 = f(y)\]
The times 1 there is just to show that I took the derivative of y (since it was the variable expression on the integrand).

So given that g'(y) = f(y), we now look at your expression for f(x).
\[f(x) = \int\limits_{0}^{sinx}\sqrt{1+t^{2}}dt \implies f(y) = \int\limits_{0}^{siny}\sqrt{1+t^{2}}dt = g'(y)\]
So if we want g''(y), we'll now differentiate that integral expression in the same way as the first:
\[g''(y) = f'(y) = \frac{ d }{ dy }\int\limits_{0}^{siny}\sqrt{1+t^{2}}dt = \sqrt{1+\sin^{2}y} * cosy = g''(y)\] So just like before, I exchanged the variable we had in our integral with the variable expression on the integrand and multiplied it by the derivative of the expression on the integrand, giving us this. So now we just want g''(pi/6), so we plug pi/6 into what we have:
\[\sqrt{1 + \sin^{2}(\pi/6)} * \cos(\pi/6) = \sqrt{ 1 + \frac{ 1 }{ 4 }}*\frac{ \sqrt{3} }{ 2 } \]
\[= \sqrt{\frac{ 4 }{ 4 }+\frac{ 1 }{ 4 }}*\frac{ \sqrt{3} }{ 2 } = \sqrt{\frac{ 5 }{ 4 }}*\frac{ \sqrt{3} }{ 2 } = \frac{ \sqrt{5} }{ 2 }*\frac{ \sqrt{3} }{ 2 } = \frac{ \sqrt{15} }{ 4 }\]