# Thread: Fundamental Theorem of Calculus help!

1. ## Fundamental Theorem of Calculus help!

So, my problem is that I know we have to replace the variable in the function with the upper limit(x), and then multiply by the derivative of x. However, the problem I have here has the lower limit of cosx, and an upper limit of sinx. I know that in order to replace our variable, the upper limit has to be unknown (an x). How would I do this?

2. ## Re: Fundamental Theorem of Calculus help! Originally Posted by scorks So, my problem is that I know we have to replace the variable in the function with the upper limit(x), and then multiply by the derivative of x. However, the problem I have here has the lower limit of cosx, and an upper limit of sinx. I know that in order to replace our variable, the upper limit has to be unknown (an x). How would I do this?
If each of $\displaystyle f~\&~g$ is a differentiable function and $\displaystyle H(x) = \int_{g(x)}^{f(x)} {h(t)dt}$ then
$\displaystyle H'(x) = \left[ {h \circ f(x)\,} \right]f'(x) - \left[ {h \circ g(x)} \right]\,g'(x)$

3. ## Re: Fundamental Theorem of Calculus help! Originally Posted by Plato If each of $\displaystyle f~\&~g$ is a differentiable function and $\displaystyle H(x) = \int_{g(x)}^{f(x)} {h(t)dt}$ then
$\displaystyle H'(x) = \left[ {h \circ f(x)\,} \right]f'(x) - \left[ {h \circ g(x)} \right]\,g'(x)$
Ah, okay. So my question is: I filled in what you told me, but I'm not sure where to go from there...

4. ## Re: Fundamental Theorem of Calculus help! Originally Posted by scorks Ah, okay. So my question is: I filled in what you told me, but I'm not sure where to go from there...
There is nothing more to do. What did you get?

6. ## Re: Fundamental Theorem of Calculus help!

I have absolutely no idea what you did. The answer is:
$\displaystyle \cos(x)[\sin^2(x)+2\sin(x)]+\sin(x)[\cos^2(x)+2\cos(x)]$

7. ## Re: Fundamental Theorem of Calculus help! Originally Posted by Plato I have absolutely no idea what you did. The answer is:
$\displaystyle \cos(x)[\sin^2(x)+2\sin(x)]+\sin(x)[\cos^2(x)+2\cos(x)]$
I multiplied cos(x) by the sin^2(x) and the 2sin(x), then the same type of thing for my sin(x). Then I took out the common factor. Also, shouldn't that be a -cos(x), since the integral of sin(x) is -cos(x)?

8. ## Re: Fundamental Theorem of Calculus help!

Just in case a picture helps... ... where (key in spoiler) ...

Spoiler: ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule). And, is the FTC: Fundamental theorem of calculus - Wikipedia, the free encyclopedia

Full size

__________________________________________________ __________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

9. ## Re: Fundamental Theorem of Calculus help!

Gotcha. Thanks! I'll try again, I think is was accidentally multiplying by the integral rather than the derivative.

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