Question

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

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

Answer 1

Answer: (D)

If f(x) is a continuous function defined on [a, b], then
$m(b-a)\le {\int }_a^{b}f(x)dx\le M(b-a)$
where, M and m are maximum and minimum values
respectively of f(x) in [a, b].
Here, f(x) = 1 + e${}^{-x^2}$ is continuous in [0,1],
Now, $0<x<1\Rightarrow x^2<x\Rightarrow e^{x^2}<e^x{e}^{-x^2}>e^{-x}$
Again, 0 < x < 1$$\begin{gathered} \Rightarrow x^2>0\Rightarrow \\ {e}^{x^2}>e^0\Rightarrow e^{-x^3}<1 \end{gathered}$$ $\therefore e^{-x}<e^{-x^2}<1,\forall x\in [0,1]$
$\Rightarrow 1+e^{-x}<1+e^{-x^2}<2,\forall x\in [0,1]$
$\Rightarrow {\int }_0^1(1+e^{-x})dx<{\int }_0^1(1+e^{-x^2})dx<{\int }_0^{1}2dx$
$\Rightarrow 2-\frac{1}{e}<{\int }_0^1(1+e^{-x^2})dx<2$