The area of the region enclosed by the curves | vedclass

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

Question

(B) Given curves are $y = x$,$y = \frac{1}{x}$,$x = e$ and the positive $X$-axis $(y = 0)$.The intersection point of $y = x$ and $y = \frac{1}{x}$ is

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(B) Given curves are $y = x$,$y = \frac{1}{x}$,$x = e$ and the positive $X$-axis $(y = 0)$.The intersection point of $y = x$ and $y = \frac{1}{x}$ is found by setting $x = \frac{1}{x}$,which gives $x^2 = 1$. Since we are in the first quadrant,$x = 1$. Thus,the point of intersection is $(1, 1)$.The region is bounded by $y = x$ from $x = 0$ to $x = 1$,and by $y = \frac{1}{x}$ from $x = 1$ to $x = e$.Required Area = $\int_{0}^{1} x \, dx + \int_{1}^{e} \frac{1}{x} \, dx$$= \left[ \frac{x^2}{2} ight]_{0}^{1} + [\ln |x|]_{1}^{e}$$= \left( \frac{1}{2} - 0 ight) + (\ln e - \ln 1)$$= \frac{1}{2} + (1 - 0) = \frac{1}{2} + 1 = \frac{3}{2} 	ext{ sq units}$.

The area of the region enclosed by the curves $y = x$,$y = \frac{1}{x}$,$x = e$ and the positive $X$-axis is

Similar Questions

If the area bounded by the curve $2y^2 = 3x$,the lines $x+y=3$,$y=0$ and outside the circle $(x-3)^2 + y^2 = 2$ is $A$,then $4(\pi + 4A)$ is equal to $.........$.

The area of the region bounded by the curves $x^{2}+y^{2}=8$ and $y^{2}=2x$ is

The area (in $sq. \, units$) of the region bounded by the curves $x^{2}+2y-1=0$,$y^{2}+4x-4=0$,and $y^{2}-4x-4=0$ in the upper half plane is $....$

The area (in sq. units) bounded by the curve $y=2x-x^2$ and the line $y=-x$ is

The area of the region $\{(x, y) \in R^2: y \geq \sqrt{|x+3|}, 5y \leq x+9 \leq 15\}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module