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

Let $R$ denote the set of all real numbers. Then the area of the region $\{(x, y) \in R \times R : x > 0, y > \frac{1}{x}, 5x - 4y - 1 > 0, 4x + 4y - 17 < 0\}$ is

The area (in sq. units) of the region $\{(x, y) : x \geq 0, x+y \leq 3, x^2 \leq 4y \text{ and } y \leq 1+\sqrt{x}\}$ is

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

The area (in sq. units) of the region $A = \{(x,y) \in R \times R | 0 \le x \le 3, 0 \le y \le 4, y \le x^2 + 3x\}$ is

The parabola $y^2 = x$ divides the circle $x^2 + y^2 = 2$ into two parts whose areas are in the ratio

Vedclass Test Series

Exam Paper Generator

Online Exam Module