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Mathematics 7 Online
OpenStudy (anonymous):

Find the derivative of the following function F(x)=the integral from x^5 to x^3 of (2t-1)^3 dt I will write it in a formula form below in the next box

OpenStudy (anonymous):

\[F(x)=\int\limits_{x^5}^{x^3}(2t-1)^3 dt\]

OpenStudy (anonymous):

This implements the fundamental theorem of calculus.\[\frac{ d }{ dx } \int\limits_{0}^{f(x)}g(t) dt=g(f(x)*f ^{'}(x)\]

OpenStudy (anonymous):

\[F(x)=\int\limits_{x5}^{x^3}(2t−1)^3dt=\int\limits_{0}^{x^3}(2t−1)^3dt - \int\limits_{0}^{x^5}(2t−1)^3dt \] \[f(x)=3(2x^3-1)^3x^2-5(2x^5-1)^3x^4\]

OpenStudy (anonymous):

The trick the the FTC is that the lower limit must be equal to zero. If it isn't the integral must be separated into two separate integrals as above.

OpenStudy (anonymous):

Thank you sooooooo much for your help!

OpenStudy (anonymous):

Is there anything you are confused about or would like more help with? and of course, you are quite welcome

OpenStudy (anonymous):

I think I am good but I guess I will find out as I progress through my homework LOL

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