The area bounded by the curve x=log (|y|),the | vedclass

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

Question

(D) Given the curve is $x = \log |y|$.This can be rewritten as $|y| = e^x$,which implies $y = \pm e^x$.The curve is symmetric about the $x$-axis.The a

Vedclass · Accepted Answer

$2\left(1-e^{-1}\right)$

Vedclass · Answer

(D) Given the curve is $x = \log |y|$.This can be rewritten as $|y| = e^x$,which implies $y = \pm e^x$.The curve is symmetric about the $x$-axis.The area bounded by the curve,the lines $x = -1$ and $x = 0$ is given by the integral of $y$ with respect to $x$ from $-1$ to $0$.Since the curve is symmetric,the total area is twice the area above the $x$-axis:$	ext{Area} = 2 \int_{-1}^{0} |y| \, dx = 2 \int_{-1}^{0} e^x \, dx$Evaluating the integral:$	ext{Area} = 2 \left[ e^x ight]_{-1}^{0}$$= 2 (e^0 - e^{-1})$$= 2 (1 - e^{-1})$Thus,the required area is $2(1 - e^{-1})$ square units.

The area bounded by the curve $x=\log (|y|)$,the lines $x=-1$ and $x=0$ is

Similar Questions

The area of the region bounded by the line $y = 3 - x$,the $X$-axis,and the lines $x = 2$ and $x = 5$ is . . . . . . .

The area enclosed between the curve $y = \log_e(x + e)$ and the coordinate axes is

Area of the region bounded by the curve $y^{2}=4x$,$y$-axis and the line $y=3$ is

The slope of the tangent to a curve $y=f(x)$ at $(x, f(x))$ is $2x+1$. If the curve passes through the point $(1,2)$,then the area (in sq. units),bounded by the curve,the $X$-axis and the line $x=1$,is

The area bounded by the curve $y = x$,the $x$-axis,and the ordinates $x = -1$ to $x = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module