Area enclosed by the graph of y = ln ( x +1), y-axis, and the line y =1 is

Question

Area enclosed by the graph of y = ln ( x +1), y-axis, and the line y =1 is (A) e -2 (B) e -1 (C) e (D) 2 e

Tardigrade Admin · Accepted Answer

.A=∫ limits01(ey-1) d y=ey-y]01 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/dc0b880d5d17c6591a1110ea266f4d8f-.png" /> = (e - 1) - (1 - 0) = e - 2

Q. Area enclosed by the graph of $y = ln (x + 1)$ , $y$ -axis, and the line $y = 1$ is

$e - 2$

$e - 1$

$e$

$2 e$

$A = 0 \int 1 (e^{y} - 1) d y = e^{y} - y]_{0}^{1}$

$= (e - 1) - (1 - 0)$
$= e - 2$

Q. Area enclosed by the graph of y=ln(x+1), y-axis, and the line y=1 is

e−2

e−1

e

2e

A=0∫1​(ey−1)dy=ey−y]01​ =(e−1)−(1−0) =e−2

Q. Area enclosed by the graph of $y = ln (x + 1)$ , $y$ -axis, and the line $y = 1$ is

$e - 2$

$e - 1$

$e$

$2 e$

$A = 0 \int 1 (e^{y} - 1) d y = e^{y} - y]_{0}^{1}$

$= (e - 1) - (1 - 0)$
$= e - 2$