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]#?

1 Answer
May 17, 2016

Answer:

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)#.