ISI2014-DCG-20

56 views

If $A(t)$ is the area of the region bounded by the curve $y=e^{-\mid x \mid}$ and the portion of the $x$-axis between $-t$ and $t$, then $\underset{t \to \infty}{\lim} A(t)$ equals

1. $0$
2. $1$
3. $2$
4. $4$
in Geometry
retagged

1 vote

According to the question,
$\displaystyle A(t)=\int_{-t}^{t} e^{-|x|}\mathrm{d}x$

Since $|x|=\left\{\begin{matrix} -x & ; x<0\\ x & ; x\ge0 \end{matrix}\right.$

Hence $\displaystyle \int_{0}^{t} e^{-|x|}\mathrm{d}x=\int_{0}^{t} e^{-x}\mathrm{d}x \tag{i}$

and

$\displaystyle \int_{-t}^{0} e^{-|x|}\mathrm{d}x=\int_{-t}^{0} e^{-(-x)}\mathrm{d}x=\int_{-t}^{0} e^{x}\mathrm{d}x \tag{ii}$

Here $x \in [-t,0)\Rightarrow x<0$.

Let $x=-y\Rightarrow \mathrm{d}x=-\mathrm{d}y$.

When $x=-t$, then $y=t$ and when $x=0$, then $y=0$

\begin{align} \displaystyle \therefore \int_{-t}^{0} e^{x}\mathrm{d}x&=\int_{t}^{0} e^{-y}(-\mathrm{d}y)\\ \displaystyle &=-\int_{t}^{0} e^{-y}\mathrm{d}y=\int_{0}^{t} e^{-y}\mathrm{d}y ~;~ \left[\because \int_{a}^{b}f(x)\mathrm{d}x=- \int_{b}^{a}f(x)\mathrm{d}x \right] \tag{iii} \end{align}

Look at the integral $\displaystyle \int_{0}^{t} e^{-y}\mathrm{d}y$. Here $y$ is just a variable. If we write $\displaystyle \int_{0}^{t} e^{-y}\mathrm{d}y= \displaystyle \int_{0}^{t} e^{-v}\mathrm{d}v$. It's also valid. Even so we can write $\displaystyle \int_{0}^{t} e^{-y}\mathrm{d}y= \displaystyle \int_{0}^{t} e^{-x}\mathrm{d}x$.

\begin{align} \therefore \int_{-t}^{0} e^{x}\mathrm{d}x &=\int_{0}^{t} e^{-x}\mathrm{d}x ~;~[\mathrm{From~no(iii)}] \\ \Rightarrow \int_{-t}^{0} e^{-|x|}\mathrm{d}x &=\int_{0}^{t} e^{-x}\mathrm{d}x ~;~[\mathrm{From~no(ii)}] \tag{iv} \end{align}

\begin{align} \therefore A(t)=\int_{-t}^{t} e^{-|x|}\mathrm{d}x&=\int_{-t}^{0} e^{-|x|}\mathrm{d}x+\int_{0}^{t} e^{-|x|}\mathrm{d}x\\ &=\int_{0}^{t} e^{-x}\mathrm{d}x+\int_{0}^{t} e^{-x}\mathrm{d}x ~;~~[\mathrm{Using~no(iv)~and~no(i)}] \\ &=2\int_{0}^{t} e^{-x}\mathrm{d}x \\ &=-2\left[ e^{-x}\right]_{0}^{t}\\ &=-2\left( e^{-t}-e^0\right)=2 \left( 1-\frac{1}{e^t} \right) \end{align}

$$\because t \to \infty \Rightarrow \frac{1}{e^t}\to 0 \\ \displaystyle \therefore \lim_{t \to \infty} A(t) = \lim_{t \to \infty} 2 \left( 1-\frac{1}{e^t} \right)= 2$$.

Shortcut Idea: It's easily intuitive that $y$-axis cuts $e^{-|x|}$ graph into half. See it below. Then evaluating the area of the right side ($x\ge0$) of the graph i.e. $\displaystyle \int_{0}^{t}e^{-x}\mathrm{d}x$ is just enough.

So, $A(t)$ will be $\displaystyle 2\times \int_{0}^{t}e^{-x}\mathrm{d}x=2\left(1-\frac{1}{e^t} \right)$

So the correct answer is C.

edited

Related questions

1
156 views
The integral $\int _0^{\frac{\pi}{2}} \frac{\sin^{50} x}{\sin^{50}x +\cos^{50}x} dx$ equals $\frac{3 \pi}{4}$ $\frac{\pi}{3}$ $\frac{\pi}{4}$ none of these
For real $\alpha$, the value of $\int_{\alpha}^{\alpha+1} [x]dx$, where $[x]$ denotes the largest integer less than or equal to $x$, is $\alpha$ $[\alpha]$ $1$ $\dfrac{[\alpha] + [\alpha +1]}{2}$
The value of the definite integral $\int_0^{\pi} \mid \frac{1}{2} + \cos x \mid dx$ is $\frac{\pi}{6} + \sqrt{3}$ $\frac{\pi}{6} - \sqrt{3}$ $0$ $\frac{1}{2}$
The value of the integral $\displaystyle{}\int_{-1}^1 \dfrac{x^2}{1+x^2} \sin x \sin 3x \sin 5x dx$ is $0$ $\frac{1}{2}$ $– \frac{1}{2}$ $1$