The area of the region R={(x,y):xy le 8, 1 le | vedclass

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

Question

(C) The region $R$ is bounded by $y=x^2$, $y=1$, and $xy=8$ (or $y=8/x$).First, find the intersection points:$x^2=1 \implies x=1$ (since $x \ge 0$)$x^

Vedclass · Accepted Answer

$\frac{2}{3}(24\,\log_{e}(2)-7)$

Vedclass · Answer

(C) The region $R$ is bounded by $y=x^2$, $y=1$, and $xy=8$ (or $y=8/x$).First, find the intersection points:$x^2=1 \implies x=1$ (since $x \ge 0$)$x^2=8/x \implies x^3=8 \implies x=2$$8/x=1 \implies x=8$The area $A$ is given by the sum of two integrals:$A = \int_{1}^{2} (x^2 - 1) dx + \int_{2}^{8} (\frac{8}{x} - 1) dx$Evaluating the first integral:$\int_{1}^{2} (x^2 - 1) dx = [\frac{x^3}{3} - x]_{1}^{2} = (\frac{8}{3} - 2) - (\frac{1}{3} - 1) = \frac{2}{3} - (-\frac{2}{3}) = \frac{4}{3}$Evaluating the second integral:$\int_{2}^{8} (\frac{8}{x} - 1) dx = [8 \ln|x| - x]_{2}^{8} = (8 \ln 8 - 8) - (8 \ln 2 - 2) = 8(3 \ln 2) - 8 - 8 \ln 2 + 2 = 24 \ln 2 - 8 \ln 2 - 6 = 16 \ln 2 - 6$Total area $A = \frac{4}{3} + 16 \ln 2 - 6 = 16 \ln 2 - \frac{14}{3} = \frac{48 \ln 2 - 14}{3} = \frac{2}{3}(24 \ln 2 - 7)$.

The area of the region $R=\{(x,y):xy \le 8, 1 \le y \le x^{2}, x \ge 0\}$ is

Similar Questions

Let $S$ be the region bounded by the curves $y=x^{3}$ and $y^{2}=x$. The curve $y=2|x|$ divides $S$ into two regions of areas $R_{1}$ and $R_{2}$. If $\max \{R_{1}, R_{2}\}=R_{2}$,then $\frac{R_{2}}{R_{1}}$ is equal to

The area bounded by the curve $y=\cos x$,the line joining $(-\pi / 4, \cos (-\pi / 4))$ and $(0,2)$ and the line joining $(\pi / 4, \cos (\pi / 4))$ and $(0,2)$ is

The area bounded by the curves $y^2 = 4x$ and $x^2 = 4y$ is

The area enclosed by the curves $y=\ln(x+e^{2})$,$x=\ln(2/y)$ (which is $y=2e^{-x}$) and $x=\ln 2$,above the line $y=1$ is:

Area lying in the first quadrant between the curves $x^2 + y^2 = \pi^2$ and $y = \sin x$ is equal to :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module