Question

The maximum possible area bounded by the parabola $y=x^2+x+10$ and a chord of the parabola of length $1$ is.

• A. $\frac{1}{12}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{1}{6}$


From the figure the maximum area
$=\underset{-1}{\overset{0}{\int }}10-(x^2+x+10)dx=-\underset{-1}{\overset{0}{\int }}(x^2+x)dx=-[\frac{x^3}{3}+\frac{x^2}{2}{]}_{-1}^0=-[\frac{1}{3}-\frac{1}{2}]=\frac{1}{6}$