The area enclosed between the curve y = log_e | vedclass

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

Question

(C) The curve is $y = \log_e(x + e)$.To find the $x$-intercept,set $y = 0$:$0 = \log_e(x + e) \implies x + e = e^0 = 1 \implies x = 1 - e$.To find the

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The curve is $y = \log_e(x + e)$.To find the $x$-intercept,set $y = 0$:$0 = \log_e(x + e) \implies x + e = e^0 = 1 \implies x = 1 - e$.To find the $y$-intercept,set $x = 0$:$y = \log_e(0 + e) = \log_e(e) = 1$.The area enclosed by the curve and the coordinate axes is given by the integral of $y$ with respect to $x$ from $x = 1 - e$ to $x = 0$:$	ext{Area} = \int_{1 - e}^0 \log_e(x + e) \, dx$.Let $t = x + e$,then $dt = dx$. When $x = 1 - e$,$t = 1$. When $x = 0$,$t = e$.$	ext{Area} = \int_1^e \log_e(t) \, dt$.Using the integration by parts formula $\int \log_e(t) \, dt = t \log_e(t) - t$:$	ext{Area} = [t \log_e(t) - t]_1^e = (e \log_e(e) - e) - (1 \log_e(1) - 1) = (e - e) - (0 - 1) = 1$.Thus,the area is $1 	ext{ sq. unit}$.

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

Similar Questions

Area of the region bounded by the curve $|x| + y = 1$ is . . . . . . sq. units.

Area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is . . . . . . .

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

Find the area of the region bounded by the parabola $y=x^{2}$ and $y=|x|$.

The area of the region bounded by the curve $y = \cos x$,the $X$-axis,and the lines $x = 0$ and $x = \pi$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module