Question

The value of ${\int }_0^1(1+e^{-x^2})dx$ is

• A. $-1$
• B. $2$
• C. $1+e^{-1}$
• D. none of these

Answer 1

The correct option is D none of these

Indefinite integral of $e^{-x^2}$ can not be expressed in terms of elementary functions. So, we will verify the options one by one.

$e^{-x^2}$ is always greater than zero.

Hence, $x$ is between $0$ and $1$, graph of $1+e^{-x^2}$ will be always above x-axis.

Hence, area under the curve when $x$ varies from $0$ to $1$ is a positive number. Option A is not possible.

Similalrly, $e^{-x^2}<1$

$\Rightarrow 1+e^{-x^2}<2$

$\Rightarrow {\int }_0^1(1+e^{-x^2})dx<{\int }_0^{1}2dx\Rightarrow {\int }_0^1(1+e^{-x^2})dx<2$

Hence, option B is not correct.

Similarly, when 'x' is between $0$ and $1$, $e^{-x}<e^{-x^2}$

$\Rightarrow {\int }_0^1(1+e^{-x})dx<{\int }_0^1(1+e^{-x^2})dx\Rightarrow {[x-e^{-x}]}_0^1<{\int }_0^1(1+e^{-x^2})dx\Rightarrow 2-e^{-1}<{\int }_0^1(1+e^{-x^2})dx$

It can be shown that $1+e^{-1}<2-e^{-1}$

Hence, $1+e^{-1}<{\int }_0^1(1+e^{-x^2})dx$

Hence, option C is also not correct.