Question

How do you use the second fundamental theorem of Calculus to find the derivative of given `int (cos(t^2)+t) dt` from `[-5, sinx]`?

Answer 1

By the FTC and the Chain Rule, the derivative is `(cos(sin^{2}(x))+sin(x)) * cos(x)`.

Explanation:

The "second" Fundamental Theorem of Calculus (I just call it part of the FTC) says `d/dx(int_{a}^{x}f(t)\ dt)=f(x)` when `f` is continuous on the interval between `a` and `x`.

Let `F(x)=int_{-5}^{x}(cos(t^{2})+t)\ dt` and `g(x)=sin(x)`. The given function for this question is `h(x)=F(g(x))`. The Chain Rule implies that `h'(x)=F'(g(x)) * g'(x)`. Since `F'(x)=cos(x^{2})+x` and `g'(x)=cos(x)`, it follows that

`h'(x)=d/dx(int_{-5}^{sin(x)}(cos(t^{2})+t)\ dt)`

`=(cos(sin^{2}(x))+sin(x)) * cos(x)`.