$$h ( x ) = \int _ { 1 } ^ { x } t ( t ) d t$$
a) $h ( 1 ) = \int ^ { \prime } f ( t ) d t = 0$
b) $$\begin{aligned} & h h ^ { \prime } ( x ) = \frac { d } { d x } \int _ { 0 } ^ { x } f ( t ) d t = f ( x ) \\ & h ^ { \prime } ( 4 ) = f ( 4 ) = 2 \end{aligned}$$
oncave up on $( 1,3 )$ and $( 6,7 )$ since $h ^ { \prime } ( x )$ is increa. those intervals
d) h(x) has no relotive minimum on [1,7] since $h ^ { \prime } ( x )$ does
change sign from - ve to tre
$\therefore$ Minimum must accur at either enopoint
$$\begin{aligned} & h ( 1 ) = 0 \\ & h ( 7 ) = \int _ { 1 } ^ { 7 } f ( t ) d t \end{aligned}$$
$h ( 7 ) > h ( 1 ) \quad$ since $h ( x )$ has a relative maximum $a t \quad x = 5$, and connot decrease to zero, $\int ^ { 5 } h ^ { \prime } ( x ) > \int _ { 5 } ^ { 7 } h ^ { \prime } ( x ) d x$