Calculus I, Section 5.3, #12
                  The Fundamental Theorem of Calculus
                                                                                                          1
                  Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
                        R(y) = Z 2t3sin(t) dt
                                 y
                  Let’s remind ourselves of the Fundamental Theorem of Calculus, Part 1:
                          The Fundamental Theorem of Calculus, Part 1 If f is continuous on [a;b], then the
                          function g defined by
                                g(x) = Z xf(t) dt     a ≤ x ≤ b
                                        a
                          is continuous on [a;b] and differentiable on (a;b) and g′(x) = f(x).
                  First, we’ll use properties of the definite integral to make the integral match the form in the Fundamental
                  Theorem.
                        Z 2t3sin(t) dt = −Z y t3sin(t) dt
                         y                   2
                  so we have
                                  R(y) = −Z yt3sin(t) dt
                                             2
                  The minus sign is just a constant factor, so
                              d [R(y)] = −1· d Z yt3sin(t) dt
                              dy              dy   2
                                 R′(y) = −1·y3sin(y)
                  Thus,
                        R′(y) = −y3sin(y)
                     1Stewart, Calculus, Early Transcendentals, p. 399, #12.