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]#?

Question

Luke Alvarez · Accepted Answer

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

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)#.

Samantha Adkison · Answer

undefined To find the derivative of the integral $\int_{-5}^{\sin(x)} (\cos(t^2) + t) \, dt$, you can use the second fundamental theorem of calculus. According to this theorem, if $F(x)$ is an antiderivative of $f(x)$, then $\frac{d}{dx} \left[\int_{a}^{g(x)} f(t) \, dt\right] = f(g(x)) \cdot g'(x)$.

First, find an antiderivative of $\cos(t^2) + t$. Let $F(t)$ be such an antiderivative. Then, compute $F(\sin(x))$ and differentiate it with respect to $x$. This will give you the derivative of the given integral.