Browse Questions

Find the area enclosed by the curve $y^2+1=x,x\leq 2$ and the line $x=2$.

$\begin {array} {1 1} (A)\;\large\frac{2}{3}\: sq\: units & \quad (B)\;\large\frac{5}{3}\: sq\: units \\ (C)\;\large\frac{3}{4}\: sq\: units & \quad (D)\;\large\frac{4}{3}\: sq\: units \end {array}$

Toolbox:
• Area of the region bounded between a curve and a line is given by $A=\int_a^by\;dx=\int_a^bf(x)\;dx$
• where a and b are the point of intersection of the line and the curve.
• The point of intersection can be found by solving the two equations.
The area of the required region is the shaded part of the curve below $y^2 + 1 = x$, intercepted by the line $x=2$
This is the shaded portion shown in the fig.
clearly the point of intersection is (1,0) with the x-axis and (2,1) and (2,-1)
Area of the region bounded between a curve and a line is given by $A=\large \int_a^b$$y\;dx=\large \int_1^2$$y \; dx$
$A = 2 \times \large \int_1^2$$\sqrt (x^2-1)\;dx A = 2 \Large \frac{[(x-1)^{3/2}]_1^2}{3/2} Applying Limits, A = 2 \times \large \frac{2}{3}$$ \times (1)^{3/2}$
$\Rightarrow A = \large\frac{4}{3}$ sq units