Ok I know the first fundamental theorem of Calc but this question is confusing me because both of the bounds are varibles with exponents. Please help

Find the derivative of https://webwork.math.ohio-state.edu/...5a7bc820c1.png

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- Jan 17th 2011, 10:54 AMrawkstar1st Fundamental Theorem of Calc
Ok I know the first fundamental theorem of Calc but this question is confusing me because both of the bounds are varibles with exponents. Please help

Find the derivative of https://webwork.math.ohio-state.edu/...5a7bc820c1.png - Jan 17th 2011, 11:07 AMFernandoRevilla
$\displaystyle F'(x)=(2x^7-1)^3\cdot 7x^6-(2x^5-1)^3\cdot 5x^4$

Fernando Revilla - Jan 18th 2011, 08:00 AMHallsofIvy
You can fix the problem that the lower limit of integration is a variable by breaking it into to integrals:

$\displaystyle \int_{x^5}^{x^7}(2t- 1)^3 dt= \int_a^{x^7}(2t- 1)^3 dt- \int_a^{x^5} (2t- 1)dt$

where "a" is any fixed number- the lower limit won't be relevant to the derivative.

To fix the problem that the upper limit is not just "x", make a change of variable. If we let $\displaystyle u= t^{1/7}$, then when [tex]t= x^7[tex], u will be equal to x. $\displaystyle t= u^7$, of course, and $\displaystyle dt= 7t^6 dt$ so the first integral becomes

$\displaystyle 7\int_{a^{1/7}}^x(2u^7- 1)^3t^6 du$.

For the second integral let $\displaystyle v= t^{1/5}$ so that when $\displaystyle t= x^5$, $\displaystyle v= x$. $\displaystyle t= v^5$ and $\displaystyle dt= 5v^4dv$ so the integral becomes

$\displaystyle 5\int_{a^{1/5}}^x (2v^5- 1)^3t^4dt$

That is, $\displaystyle \int_{x^5}^{x^7}(2x-1)^2dt= 7\int_{a^{1/7}}^x (2u^7- 1)^3u^6du- 5\int_{a^{1/5}}^x (2v^5- 1)v^4 dv$

and now you can easily apply the Fundamental Theorem of Calculus to the two integrals on the right. The derivative with respect to x is, modulo any careless errors,

$\displaystyle 7(2x- 1)^3x^6- 5(2x^5-1)x^4$

In fact, that can be generalized to "Leibniz' formula":

$\displaystyle \frac{d}{dx}\left(\int_{\phi(x)}^{\psi(x)} f(x,t)dt= f(x, \psi(x))\frac{d\psi(x)}{dx}- f(x,\phi(x))\frac{d\phi(x)}{dx}+ \int_{\phi(x)}^{\psi(x)}\frac{\partial f(x,t)}{\partial x}dt$

which is, I suspect, how FernandoRevilla got his answer so quickly! - Jan 18th 2011, 08:33 AMFernandoRevilla
I got it in the following way:

$\displaystyle F(x)=\displaystyle\int_{g(x)}^{h(x)}f(t)dt=\displa ystyle\int_{0}^{h(x)}f(t)dt-\displatytyle\int_{0}^{g(x)}f(t)dt$

Using the Fundamental Theorem of Calculus and Chain's Rule we inmediately obtain:

$\displaystyle F'(x)=f(h(x))h'(x)-f(g(x))g'(x)$

(of course $\displaystyle g,h$ differentiable, etc)

Fernando Revilla