- #1
marathon
- 3
- 0
i have a similar one.
f(x) = \int\frac{dt}{\sqrt{1+t^3}} on (0, g(x))
g(x) = \int(1+sin(t^2))dt on (0, cos(x))
that is, these are definite integrals on the interval from zero up to the given function.
the question is to solve f'(pi/2). the correct answer is -1 but i don't understand how to get it. I'm aware that integrating sin(t^2) leads to a fresnel, but I'm pretty sure that's not what we're supposed to be doing here since it's a challenge problem from a calc 2 text on the chapter with FTC. what i got was that g(pi/2) should be 0 since cos(pi/2)=0 and any integral on (a,a)=0. then we'd have f'(pi/2) = \frac{1}{\sqrt{1+g(\frac{\pi}{2})^3}}=\frac{1}{\sqrt{1+0^3}} = 1, but obviously this is incorrect. halp?
f(x) = \int\frac{dt}{\sqrt{1+t^3}} on (0, g(x))
g(x) = \int(1+sin(t^2))dt on (0, cos(x))
that is, these are definite integrals on the interval from zero up to the given function.
the question is to solve f'(pi/2). the correct answer is -1 but i don't understand how to get it. I'm aware that integrating sin(t^2) leads to a fresnel, but I'm pretty sure that's not what we're supposed to be doing here since it's a challenge problem from a calc 2 text on the chapter with FTC. what i got was that g(pi/2) should be 0 since cos(pi/2)=0 and any integral on (a,a)=0. then we'd have f'(pi/2) = \frac{1}{\sqrt{1+g(\frac{\pi}{2})^3}}=\frac{1}{\sqrt{1+0^3}} = 1, but obviously this is incorrect. halp?