How do you use the Fundamental Theorem of Calculus to find the derivative of #int (u^3) / (1+u^2) du# from 2-3x to 5?

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Answer 1 · 2017-02-17T23:05:48

#= (3(2-3x)^3) / (1+(2-3x)^2) #

The following assumes you mean the derivative wrt #x#.

We start with the formal statement of the FTC, Part I:

#d/(du) ( int _{a}^{u} f(t) \ dt ) = f(u)# where #a =# some constant.

If in fact #u = u(x)#, we can use the Chain Rule so that:

#d/(dx) ( int _{a}^{u(x)} f(t) \ dt ) = f(u(x) ) (du)/(dx)#

We can also play with the intervals ..... in summary #int_a^b = int_a^0 + int_0^b = int_0^b - int_0^a#.... in order to reach this point:

#d/(dx) ( int _{u_1(x)}^{u_2(x)} f(t) \ dt ) = f(u_2(x) ) (du_2)/(dx) - f(u_1(x) ) (du_1)/(dx)#

Gasp! With all that, we just need to process this:

#d/dx ( int_(2-3x)^5 (u^3) / (1+u^2) du )#

#= (5^3) / (1+5^2) (d (5))/(dx) - ((2-3x)^3) / (1+(2-3x)^2) (d (2-3x))/(dx)#

#= (3(2-3x)^3) / (1+(2-3x)^2) #

Maybe you can further simplify that.