CUET Preparation Today
The area of the region bounded by the curve $y=x^2+2, y=-x, x=0$ and x=1 is : |
$\frac{5}{16}$ $\frac{7}{16}$ $\frac{17}{6}$ $\frac{6}{17}$ |
$\frac{17}{6}$ |
The correct answer is Option (3) → $\frac{17}{6}$ so area = $\int\limits_0^1x^2+2dx+\frac{1}{2}×1×1$ $=\left[\frac{x^3}{x}+2x\right]_0^1+\frac{1}{2}$ $=\frac{7}{3}+\frac{1}{2}=\frac{17}{6}$ |
