Mathematics - Area Under Curves Question with Solution | TestHub

The given curve $y=e^{-x^2}$ Draw a rough sketch of curve at $x=0,y=1$ and at $x=1,y=1/e$ $\because y=e^{-x^2}$ $\Rightarrow \frac{dy}{dx}=-2x{e}^{-x^2}<0\forall x\in (0,1)$ $\therefore y=e^{-x^2}$ is decreasing on $(0,1)$ Hence its graph is as shown in figure given below
Now, $\mathrm{S}=$ area exclosed by curve $=\mathrm{XYCO}$ and area of rectangle $\mathrm{OCYL}=\frac{1}{e}$ Clearly $S>\frac{1}{e}\therefore$ a is true. For $x\in [0,1]\Rightarrow x^2<x$ $\begin{array}{l}\Rightarrow -x^2>-x\Rightarrow e^{-x^2}\ge e^{-x}\forall x\in [0,1]\\ \Rightarrow {\int }_0^1{e}^{-x^2}dx>{\int }_0^1{e}^{-x}dx=1-\frac{1}{e}\\ \Rightarrow S>1-\frac{1}{e}\therefore \text{ (b) is true. }\end{array}$ Now $S<$ area of rectangle $\mathrm{XADO}+$ area of rectangle $\mathrm{ZDCN}$ $\Rightarrow S<\frac{1}{\sqrt{2}}\times 1+\left(1-\frac{1}{\sqrt{2}}\right)\frac{1}{\sqrt{e}}$ $\therefore S<\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)\because (\mathrm{d})$ is true Also as $\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)<1-\frac{1}{e}\therefore$ (c) is incorrect.