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?

1 Answer
Feb 17, 2017

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

Explanation:

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.