Question

Let $g(x)=\underset{0}{\overset{x}{\int }}f(t)dt$, where $f$ is a function whose graph is show adjacently.

Maximum value of $g(x)$ in $x\in [0,7]$ is -

• A. $3$
• B. $9/2$
• C. $3/2$
• D. $6$

Answer 1

Answer: (A)

$g(x)=\underset{0}{\overset{x}{\int }}f(t)dt$
$g^{\prime }(x)=f(x)$
From the graph it is clear that
$f(x)>0$ in $x\in [0,3)$ and
$f(x)<0$ in $x\in (3,7)$
$\therefore g(x)$ is increasing in $[0,3]$ and
$g(x)$ is decreasing in $[3,7]$
$\therefore$ maximum value of $g(x)$ occurs at $x=3$
$\therefore g(3)=\underset{0}{\overset{3}{\int }}f(t)dt$
$=\underset{0}{\overset{1}{\int }}1\cdot dt+\underset{1}{\overset{2}{\int }}(2t-1)dt+\underset{2}{\overset{3}{\int }}(3t+9)dt$
$=1+{\left(t^2-t\right)}_1^2+{\left(9t-3\frac{t^2}{2}\right)}_2^3$
$=1+(4-2-0)+\left(27-\frac{27}{2}-18+6\right)=\frac{9}{2}$