Question

The area enclosed between the curve $y=1+x^2$, the y-axis and the straight line $y=5$ is given by

• A. $\frac{14}{3}$ sq unit
• B. $\frac{7}{3}$ sq unit
• C. $5$ sq unit
• D. $\frac{16}{3}$ sq unit

Answer 1

The correct option is D $\frac{16}{3}$ sq unit

Given curve is $y=1+x^2$ and line $y=5$
$\therefore$ Required area $={\int }_1^{5}xdy={\int }_1^5\sqrt{y-1}dx$
$={\left[\frac{{(y-1)}^{3/2}}{3/2}\right]}_1^5=\frac{2}{3}[{\left(4\right)}^{3/2}-0]$
$=\frac{16}{3}$ sq unit.