Help please? Integral question below. Don’t really understand the concept of integrals in general. Thanks!

Question

1229 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-08T16:19:58

For $g (y) = \int_{3}^{y} f (x) d x$ $g(y) = int_3^y f(x) dx$ , FTC part 1 gives us

$g'(y) = f(y)$

Now we want $g''(y) = f'(y)$ where $f(y) = int_0^siny sqrt(1+t^2) dt$

Again, using FTC part 1, this time with the chain rule as well, we get:

$g''(y) = f'(y) = sqrt(1+sin^2y) * d/dy(siny)$

$= sqrt(1+sin^2y) cosy$

At $y=pi/6$ , we get

$g''(pi/6) = sqrt(1+sin^2(pi/6)) cos(pi/6)$

$= sqrt(1+1/4) (sqrt3/2) = sqrt15/4$