The area of the region enclosed by the curves | vedclass

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

Question

(D) The curves are $y=e^x$ and $y=|e^x-1|$.For $x < 0$,$e^x < 1$,so $|e^x-1| = 1-e^x$.The curves intersect when $e^x = 1-e^x$,which implies $2e^x = 1$

Vedclass · Accepted Answer

$1-\ln 2$

Vedclass · Answer

(D) The curves are $y=e^x$ and $y=|e^x-1|$.For $x The curves intersect when $e^x = 1-e^x$,which implies $2e^x = 1$,or $e^x = 1/2$,so $x = \ln(1/2) = -\ln 2$.The area is bounded by $x = -\ln 2$ to $x = 0$ between the curves $y=e^x$ and $y=1-e^x$.Area $= \int_{-\ln 2}^{0} [e^x - (1-e^x)] \, dx$$= \int_{-\ln 2}^{0} (2e^x - 1) \, dx$$= [2e^x - x]_{-\ln 2}^{0}$$= (2e^0 - 0) - (2e^{-\ln 2} - (-\ln 2))$$= (2 - 0) - (2(1/2) + \ln 2)$$= 2 - (1 + \ln 2)$$= 1 - \ln 2$.

The area of the region enclosed by the curves $y=e^x$,$y=|e^x-1|$ and the $y$-axis is:

Similar Questions

The area bounded by the curve $y = f(x)$,the $x$-axis,and the ordinates $x = 1$ and $x = b$ is $(b - 1) \sin(3b + 4)$. Then $f(x)$ is:

The area of the region enclosed by the lines $y = 2x$, $2y = x$ and $x = 2$ (in sq. units) is...

Area of the region bounded by $y = \cos x$,$x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$ is . . . . . . .

The area of the region bounded by the line $y=3x$ and the curve $y=x^2$ in square units is

The area bounded by the curve $y = x |\sin(\frac{x}{2})|$ and the $X$-axis between the lines $x = 0$ and $x = 4\pi$ (in square units) is: (in $\pi$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module