Question

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

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

Answer 1

The correct option is D none of these

If $f\left(x\right)$ is a continuous function defined on $[a,b]$,
then $m(b-a)\le {\int }_a^{b}f\left(x\right)dx\le M(b-a)$
Where M and m are maximum and minimum values respectively of $f\left(x\right)$ in $[a,b]$
Here, $f\left(x\right)=1+e^{-x^2}$ is continuous in $[0,1]$.
Now, $0<x<1\Rightarrow x^2<x\Rightarrow e^{x^2}<e^x\Rightarrow e^{-x^2}>e^{-x}$
Again, $0<x<1\Rightarrow x^2>0\Rightarrow e^{x^2}>e^0\Rightarrow e^{-x^2}<1$
Therefore $e^{-x}<e^{-x^2}<1$ for all $xϵ[0,1]$
$\Rightarrow 1+e^{-x}<1+e^{-x^2}<2$ for all $e^{-x}<e^{-x^2}xϵ[0,1]$
$\Rightarrow {\int }_0^1(1+e^{-x})dx<{\int }_0^1(1+e^{-x^2})dx<{\int }_0^{1}2dx\Rightarrow 2-\frac{1}{e}<{\int }_0^1(1+e^{-x^2})dx<2$