If the area of the region bounded by the curv | vedclass

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

Question

(B) The intersection point of the curves $y = x^2$ and $y = \frac{1}{x}$ is found by setting $x^2 = \frac{1}{x}$,which gives $x^3 = 1$,so $x = 1$. Thu

Vedclass · Accepted Answer

$e^{2/3}$

Vedclass · Answer

(B) The intersection point of the curves $y = x^2$ and $y = \frac{1}{x}$ is found by setting $x^2 = \frac{1}{x}$,which gives $x^3 = 1$,so $x = 1$. Thus,the intersection point is $(1, 1)$.The area of the region bounded by the curves $y = x^2$,$y = \frac{1}{x}$,the $x$-axis $(y = 0)$,and the line $x = t$ $(t > 1)$ is given by the sum of two integrals:$	ext{Area} = \int_0^1 x^2 \, dx + \int_1^t \frac{1}{x} \, dx$Evaluating the integrals:$	ext{Area} = \left[ \frac{x^3}{3} ight]_0^1 + \left[ \ln(x) ight]_1^t$$	ext{Area} = \left( \frac{1}{3} - 0 ight) + (\ln(t) - \ln(1))$Since $\ln(1) = 0$,we have:$	ext{Area} = \frac{1}{3} + \ln(t)$Given that the area is $1 \, 	ext{sq. unit}$:$\frac{1}{3} + \ln(t) = 1$$\ln(t) = 1 - \frac{1}{3} = \frac{2}{3}$Taking the exponential of both sides:$t = e^{2/3}$

If the area of the region bounded by the curves $y = x^2$,$y = \frac{1}{x}$ and the lines $y = 0$ and $x = t$ $(t > 1)$ is $1 \, \text{sq. unit}$,then $t$ is equal to

Similar Questions

The area (in square units) bounded by the curves $y^2=4x$ and $x^2=4y$ in the plane is

The area of the region,bounded by the parabola $y=x^2+2$ and the lines $y=x, x=0$ and $x=3$,is

The area enclosed by the curves $y = \sin^{-1}(\cos x)$ and $y = \cos^{-1}(\sin x)$ for $x \in \left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]$ is:

Let the area of the region $\{(x, y): x-2y+4 \geq 0, x+2y^2 \geq 0, x+4y^2 \leq 8, y \geq 0\}$ be $\frac{m}{n}$,where $m$ and $n$ are coprime numbers. Then $m+n$ is equal to

The area of the region $S = \{(x, y) : y^{2} \leq 8x, y \geq \sqrt{2}x, x \geq 1\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module