Integrals

[intervade]

Ok, Im given a integral f(x) with a lower bound 0 and an upper bound of g(x). Its 1/(1+t^3)^(1/2)dt

g(x) is the integral from 0 to cosx of (1+sin(t^2))dt. I need to find f '(pi/2). Any suggestions on where to start? This is calc one. I can find g '(x) but I dont think thats what is required. I am not sure however on how to actually integrate it.

f '(x) would be 1/(1+(g(x))^3) but I Dont know what to do with g(x).

 

[galactus]

I have a feeling this may have to do with the second fundamental theorem of calculus.

\(\displaystyle \frac{d}{dx}\int_{a}^{g(x)}f(t)dt=f(g(x))g'(x)\)

or

\(\displaystyle \frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)\)

 

[intervade]

Yup, I agree! But my problem is g(x).. if I plug in pi/2 for g(x) I get the integral from 0 to 1 which is fine, but how do I integrate g(x) with elementary functions? Finding g'(x) is easy, but finding a value for g(x) is where Im struggling.

 

[BigGlenntheHeavy]

\(\displaystyle If \ this \ is \ true:\)

\(\displaystyle f(x) \ = \ \int_{0}^{g(x)}\frac{dt}{[1+t^{3}]^{1/2}} \ and \ g(x) \ = \ \int_{0}^{cos(x)}[1+sin(t^{2})]dt\)

\(\displaystyle Then \ g'(x) \ = \ [-sin(x)][sin(cos^{2}(x))+1] \ and \ f'(x) \ = \ \frac{g'(x)}{[1+[g(x)]^{3}]^{1/2}}\)

\(\displaystyle Ergo, \ f'(\pi/2) \ = \ \frac{g'(\pi/2)}{[1+[g(\pi/2)]^{3}]^{1/2}} \ = \ \frac{-1}{(1+0^{3})^{1/2}} \ = \ -1\)

\(\displaystyle Note: \ g(x) \ = \ \int_{0}^{cos(x)}[1+sin(t^{2})]dt, \ g(\pi/2) \ = \ \int_{0}^{cos(\pi/2)}[1+sin(t^{2})]dt,\)

\(\displaystyle g(\pi/2) \ = \ \int_{0}^{0}[1+sin(t^{2})]dt \ = \ 0, \ cos(\pi/2) \ = \ 0, \ not \ 1.\)

 

[intervade]

Ahhh haha thank you for pointing out to me that the cos(pi/2) is indeed 0. Sometimes I tend to look more into the calc rather than the simple trig or algebra!

 

[galactus]

Good, Glenn. I was not sure what was trying to be explained.