Answer
a) $g(0)=0$
$g(1)=0.5$
$g(2)=0$
$g(3)=-0.5$
$g(4)=0$
$g(5)=1.5$
$g(6)=4$
b) $g(7)=6$
c) Maximum: $t=1$
Minimum: $t=3$, $t=6$
d) See image
Work Step by Step
$f(t)$ is a piecewise function and can be written as:
$-t+1$, from $0\leq t\leq2$
$t-3$, from $2\leq t\leq6$
$-3(t-6)^2+3$, from $6\leq t\leq7$
a) $g(0)=\int_0^0f(t)dt=0$
$g(1)=\int_0^1f(t)dt=\int_0^1(-t+1)dt=[-\frac{t^2}{2}+t]_0^1=0.5$
$g(2)=\int_0^2f(t)dt=\int_0^2(-t+1)dt=[-\frac{t^2}{2}+t]_0^2=0$
$g(3)=\int_0^3f(t)dt=
\int_0^2(-t+1)dt+
\int_2^3(t-3)dt=[-\frac{t^2}{2}+t]_0^2+[\frac{t^2}{2}-3t]_2^3=-0.5$
$g(4)=\int_0^4f(t)dt=
\int_0^2(-t+1)dt+
\int_2^4(t-3)dt=[-\frac{t^2}{2}+t]_0^2+[\frac{t^2}{2}-3t]_2^4=0$
$g(5)=\int_0^5f(t)dt=
\int_0^2(-t+1)dt+
\int_2^5(t-3)dt=[-\frac{t^2}{2}+t]_0^2+[\frac{t^2}{2}-3t]_2^5=1.5$
$g(6)=\int_0^6f(t)dt=
\int_0^2(-t+1)dt+
\int_2^6(t-3)dt=[-\frac{t^2}{2}+t]_0^2+[\frac{t^2}{2}-3t]_2^6=4$
b) $g(7)=\int_0^7f(t)dt=
\int_0^2(-t+1)dt+
\int_2^6(t-3)dt+
\int_6^7(-3(t-6)^2+3)dt
=[-\frac{t^2}{2}+t]_0^2+[\frac{t^2}{2}-3t]_2^6
+[-x^3+18x^2-105x]_6^7
=6$
c) $g(t)$ has a maximum and minimum values when its derivative is 0. It has a maximum value when slope is changing from positive to negative and a minimum value when slope is changing from negative to positive. From the given graph, we can see $f(t)=0$ when $t=1$, $t=3$, and $t=6$. $t=1$ changes from positive to negative, so it is a maximum and $t=3$ and $t=6$ changes from negative to positive, so it a minimum.
d) See sketch of graph $g(t)$.
