Question
Let \(g(t)=\int_{0}^{t}f(x)dx\) , and consider the graph of f shown below.
(A) Evaluate g(0), g(2), and g(6).
(B) On what interval(s) is g increasing (if any)? Justify your answer.
(C) At what value(s) of t does g have a minimum value? Justify your answer.
(D) On what interval(s) is g concave down? Justify your answer.
Answer/Explanation
(A)\(g(0)=\int_{0}^{0}f(x)dx=0\)
\(g(2)=\int_{0}^{2}f(x)dx=\int_{0}^{2}(4-4x)dx=\left [ 4x-2x^{2} \right ]^{2}_{0}=0\)
\(g(6)=\int_{0}^{6}f(x)dx=\int_{0}^{2}(4-4x)dx+\int_{3}^{2}(2x-8)dx+\int_{3}^{5}(4x-14)dx+\int_{5}^{6}6dx\)
\(=0+(-3)+4+6=7\)
(B) g is increasing on \([0,1)\cup (\frac{7}{2},6]\),since g'(t)=f(t)>0 on these intervals. Note that \(\frac{7}{2}\) is the solution to 4x -14 = 0.
(C) At t=7/2,g has a minimum value. g(0)=0, g(7/2)=-7/2, g(6)=7
(D) Since g′(t) = f(t) is decreasing only on (0, 2), you see that g′′(x)<0 on this interval. Therefore, g is concave down only on (0, 2).