The area enclosed between the curve y = log_e | vedclass

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(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

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(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 $1 	ext{ sq. unit}$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

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

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