Practicing Success
Let S be the area of the region enclosed by $y=e^{-x^2} ,y=0,x=0$ and $x=1$. Then, which of the following is incorrect? |
$S≥\frac{1}{e}$ $S≥1-\frac{1}{e}$ $S≤\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$ $S≤\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$ |
$S≤\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$ |
Clearly, $S = \int\limits_0^1e^{-x^2}dx$ Also, $⇒S>OP ×OR⇒ S>1×\frac{1}{e}⇒ S>\frac{1}{e}$ Now, $e^{-x^2}≥e^{-x}$ for all $x∈[0, 1]$ $⇒\int\limits_0^1e^{-x^2}dx≥\int\limits_0^1e^{-x}dx$ $⇒S≥[-e^{-x}]_0^1⇒S≥(-\frac{1}{e}+1)⇒S≥1-\frac{1}{e}$ Clearly, S ≤ Area of rectangle OASQ + Area of rectangle PQBU $⇒S≤\frac{1}{\sqrt{2}}×1(1-\frac{1}{\sqrt{2}})\frac{1}{\sqrt{e}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}(1-\frac{1}{\sqrt{2}})$ Since $\frac{1}{4}(1+\frac{1}{\sqrt{e}})<1-\frac{1}{e}$. Therefore, $S>\frac{1}{4}(1+\frac{1}{\sqrt{e}})$ So, option (3) is incorrect. |