Question

If $f\left(x\right)=x-x^2+1$ & $g\left(x\right)=max\left\{f\right(t);0\le t<x\}$, then $\stackrel{1}{\underset{0}{\int }}g\left(x\right)dx=?$

• A. $\frac{29}{24}$
• B. $\frac{7}{6}$
• C. $\frac{5}{4}$
• D. none of these

Answer 1

The correct option is A $\frac{29}{24}$

$f\left(x\right)=x-x^2+1$ & $g\left(x\right)=max\left\{f\right(t);0\le t<x\}$

y = g(x)

So, $\stackrel{1}{\underset{0}{\int }}g\left(x\right)dx=\stackrel{1/2}{\underset{0}{\int }}f\left(x\right)dx+[1-\frac{1}{2}]\times \frac{5}{4}$

$={[-\frac{x^3}{3}+\frac{x^2}{2}+x]}_0^{1/2}+\frac{5}{8}=\frac{29}{24}$