The area of the region given by {(x, y): xy l | vedclass

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

Question

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

Vedclass · Accepted Answer

$16 \log _{ e } 2-\frac{14}{3}$

Vedclass · Answer

(B) The region is bounded by $y = 1$,$y = x^2$,and $xy = 8$ (or $y = 8/x$).First,find the intersection points:$x^2 = 1 \implies x = 1$ (for $x > 0$).$x^2 = 8/x \implies x^3 = 8 \implies x = 2$.$8/x = 1 \implies x = 8$.The area is given by the sum of two integrals:Area $= \int \limits_1^2 (x^2 - 1) dx + \int \limits_2^8 (8/x - 1) dx$$= \left[ \frac{x^3}{3} - x \right]_1^2 + \left[ 8 \ln|x| - x \right]_2^8$$= \left( (8/3 - 2) - (1/3 - 1) \right) + \left( (8 \ln 8 - 8) - (8 \ln 2 - 2) \right)$$= (2/3 - (-2/3)) + (8(3 \ln 2) - 8 - 8 \ln 2 + 2)$$= 4/3 + 24 \ln 2 - 8 \ln 2 - 6$$= 16 \ln 2 + 4/3 - 6$$= 16 \ln 2 - 14/3$.

The area of the region given by $\{(x, y): xy \leq 8, 1 \leq y \leq x^2\}$ is :

Similar Questions

The area enclosed by the curves $y=8x-x^2$ and $8x-4y+11=0$ is

Area of the region enclosed between the curves $y^2=4(x+7)$ and $y^2=5(2-x)$ is

The area enclosed by the parabola ${y^2} = 4ax$ and the straight line $y = 2ax$ is

The area of the region bounded by the curves $x=y^2-2$ and $x=y$ is

The area bounded by the curves $|x| + |y| \geq 1$ and $x^2 + y^2 \leq 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module