Question

Let $g ( x )=\int\limits_0^{ x } f( t ) dt$, 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)

$g(x)=\int\limits_0^x f(t) d t$
$g^{\prime}(x)=f(x)$
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 $x =3$
$\therefore g (3)=\int\limits_0^3 f( t ) dt$
$=\int\limits_0^1 1 \cdot d t+\int\limits_1^2(2 t-1) d t+\int\limits_2^3(3 t+9) d t$
$=1+\left(t^2-t\right)_1^2+\left(9 t-3 \frac{t^2}{2}\right)_2^3$
$=1+(4-2-0)+\left(27-\frac{27}{2}-18+6\right)=\frac{9}{2}$