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Mathematics Question on Definite Integral

The value of the definite integral ∫01(1+e−x2)dx is\int _0^1 (1+e^{-x^2}) dx \, is

A

1

B

2

C

1+e−11+e^{-1}

D

None of these

Answer

None of these

Explanation

Solution

If f(x) is a continuous function defined on [a, b], then
             m(b−a)≤ ∫abf(x)dx≤ M(b−a)\ \ \ \ \ \ \ \ \ \ \ \ \ 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−x2^{-x^2} is continuous in [0,1],
Now, 0<x<1 ⇒  x2<x ⇒  ex2<ex  e−x2>e−x0 < x < 1 \ \Rightarrow \ \ x^2 < x \ \Rightarrow \ \ e^{x^2} < e^x \ \ e^{-x^2} > e^{-x}
Again, 0 < x < 1⇒  x2>0 ⇒ ex2>e0 ⇒ e−x3<1\Rightarrow \ \,x^2 >0 \ \Rightarrow \\\ e^{x^2} > e^0 \ \Rightarrow \ e^{-x^3} <1
∴           e−x<e−x2<1,∀ x∈ [0,1]\therefore \ \ \ \ \ \ \ \ \ \ \ e^{-x}< e^{-x^2} < 1, \forall \ x \in \ [0,1]
⇒      1+e−x<1+e−x2<2,∀x∈[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