Question
Area of the region bounded by the curve $y=e^{x}$ and lines $x=0$ and $y=e$ is [More than One Option Correct 2009](a) $e-1$(b) $\int_{1}^{\theta} \ln (e+1-y) d y$(c) $e-\int_{0}^{1} e^{x} d x$(d) $\int_{1}^{x} \ln y d y$
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This can be written as: \[A = \int_{0}^{1} e^{x} dx\] Show more…
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Area of the region bounded by the curve $y=e^{x}$ and lines $x=0$ and $y=$ $e$ is $[2009]$ (a) $e-1$ (b) $\int_{1}^{e} \ln (e+1-y) d y$ (c) $e-\int_{0}^{1} e^{x} d x$ (d) $\int_{1}^{e} \ln y d y$
Let $D$ be the region bounded by $y=1, \quad y=x$ $y=\ln x,$ and the $x$ -axis. a. Show that $\iint_{D} y d A=\int_{0}^{1} \int_{0}^{x} y d y d x+\int_{1}^{e} \int_{\ln x}^{1} y d y d x$ dividing $D$ into two regions of Type I. b. Evaluate the integral $\iint_{D} y d A$.
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The area of the region bounded by the curve $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$, where $-\infty<a<b<-2$, is [2008] (a) $\int_{a}^{b} \frac{x}{3\left((f(x))^{2}-1\right)} d x+b f(b)-a f(a)$ (b) $-\int_{a}^{b} \frac{x}{3\left((f(x))^{2}-1\right)} d x+b f(b)-a f(a)$ (c) $\int_{a}^{b} \frac{x}{3\left((f(x))^{2}-1\right)} d x-b f(b)+a f(a)$ (d) $-\int_{a}^{b} \frac{x}{3\left((f(x))^{2}-1\right)} d x-b f(b)+a f(a)$
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