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

Answer: (D)

If f(x) is a continuous function defined on [a, b], then
$ \ \ \ \ \ \ \ \ \ \ \ \ \ m(b-a) \le \ \int_a^bf(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$\Rightarrow \ \,x^2 >0 \ \Rightarrow \\ e^{x^2} > e^0 \ \Rightarrow \ e^{-x^3} <1$ $\therefore \ \ \ \ \ \ \ \ \ \ \ e^{-x}< e^{-x^2} < 1, \forall \ x \in \ [0,1]$
$\Rightarrow \ \ \ \ \ \ 1+e^{-x} < 1+e^{-x^2} < 2, \forall x \in [0,1] $
$\Rightarrow \ \ \ \ \int_0^1(1+e^{-x}) dx < \int_0^1(1+e^{-x^2}) dx <\int_0^1 2 dx$
$\Rightarrow \ \ \ \ 2-\frac{1}{e} < \int_0^1 (1+e^{-x^2}) dx <2$