The following figure shows the graph of a con | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given points are $A(1,1), B(3,2), C(2,3)$.The slope of line $l_2$ passing through $A(1,1)$ and $B(3,2)$ is $m = \frac{2-1}{3-1} = \frac{1}{2}$.Sin

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given points are $A(1,1), B(3,2), C(2,3)$.The slope of line $l_2$ passing through $A(1,1)$ and $B(3,2)$ is $m = \frac{2-1}{3-1} = \frac{1}{2}$.Since $l_1$ is parallel to $l_2$ and tangent to the curve at $C(2,3)$,the equation of line $l_1$ is $y - 3 = \frac{1}{2}(x - 2)$,which simplifies to $y = \frac{x}{2} + 2$.The area of the shaded region is the area between the line $l_1$ and the curve $f(x)$ from $x=1$ to $x=3$,minus the area of the trapezoid formed by the line $l_2$ and the $x$-axis,or more simply,the area between $l_1$ and $l_2$ minus the area between $f(x)$ and $l_2$.Alternatively,the shaded area is $\int_{1}^{3} (l_1(x) - f(x)) dx$.Area $= \int_{1}^{3} (\frac{x}{2} + 2) dx - \int_{1}^{3} f(x) dx$.Given $\int_{1}^{3} f(x) dx = 4$.Area $= [\frac{x^2}{4} + 2x]_{1}^{3} - 4$.Area $= (\frac{9}{4} + 6) - (\frac{1}{4} + 2) - 4 = (2 + 4) - 4 = 2$ sq units.

Similar Questions

The area of the region bounded by the curves $x + 3y^2 = 0$ and $x + 4y^2 = 1$ is equal to: (in $/3$)

Area of the region $\{(x, y): x^2+(y-2)^2 \leq 4, x^2 \geq 2y\}$ is

The area bounded by $y=x+1$, $y=\cos x$ and the $X$-axis is

The area of the region,enclosed by the circle $x^{2}+y^{2}=2$ which is not common to the region bounded by the parabola $y^{2}=x$ and the straight line $y=x$,is

The area of the region bounded by $y-x=2$ and $x^{2}=y$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module