Area of the region bounded by the curve y=e^x | vedclass

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(D) The curve is $y=e^x$,which implies $x=\ln y$. The region is bounded by $x=0$ (the $y$-axis),$y=e^x$,and $y=e$. The intersection of $y=e^x$ and $y=

Vedclass · Accepted Answer

$(B), (C), (D)$

Vedclass · Answer

(D) The curve is $y=e^x$,which implies $x=\ln y$. The region is bounded by $x=0$ (the $y$-axis),$y=e^x$,and $y=e$. The intersection of $y=e^x$ and $y=e$ is $e^x=e$,so $x=1$. The area $A$ can be calculated in two ways: $1$. Integrating with respect to $x$: The area is $\int_0^1 (e - e^x) dx = [ex]_0^1 - \int_0^1 e^x dx = e - \int_0^1 e^x dx$. This matches option $(C)$. Evaluating this gives $e - (e^1 - e^0) = e - e + 1 = 1$. $2$. Integrating with respect to $y$: The region spans $y$ from $1$ to $e$. The right boundary is $x=1$ and the left boundary is $x=\ln y$. So,$A = \int_1^e (1 - \ln y) dy = [y - (y \ln y - y)]_1^e = [2y - y \ln y]_1^e = (2e - e) - (2 - 0) = e - 2$. Wait,let's re-evaluate. The region is bounded by $x=0$,$y=e$,and $y=e^x$. For $y \in [1, e]$,$x$ goes from $0$ to $\ln y$. Area $= \int_1^e \ln y dy$. This matches option $(D)$. Evaluating $\int_1^e \ln y dy = [y \ln y - y]_1^e = (e \ln e - e) - (1 \ln 1 - 1) = (e - e) - (0 - 1) = 1$. Option $(A)$ is $e-1$,which is not $1$. Option $(B)$ is $\int_1^e \ln(e+1-y) dy$. Let $u = e+1-y$,then $du = -dy$. When $y=1, u=e$; when $y=e, u=1$. $\int_e^1 \ln(u) (-du) = \int_1^e \ln u du = 1$. So $(B)$ is correct. Thus,$(B), (C), (D)$ are correct.

Area of the region bounded by the curve $y=e^x$ and lines $x=0$ and $y=e$ is

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