Definite Integration Question 13
Question 13
- The value of the definite integral $\int_{0}^{1}\left(1+e^{-x^{2}}\right) d x$ is
(a) -1
(b) 2
(1981, 2M)
(c) $1+e^{-1}$
(d) None of the above
Objective Question II
(One or more than one correct option)
Show Answer
Answer:
Correct Answer: 13. (d)
Solution:
- If $f(x)$ is a continuous function defined on $[a, b]$, then
$$ m(b-a) \leq \int_{a}^{b} f(x) d x \leq M(b-a) $$
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, $0<x<1 \Rightarrow x^{2}<x \Rightarrow e^{x^{2}}<e^{x} \Rightarrow e^{-x^{2}}>e^{-x}$
Again, $0<x<1 \Rightarrow x^{2}>0 \Rightarrow e^{x^{2}}>e^{0} \Rightarrow e^{-x^{2}}<1$
$$ \begin{aligned} & \therefore \quad e^{-x}<e^{-x^{2}}<1, \forall \quad x \in[0,1] \ & \Rightarrow \quad 1+e^{-x}<1+e^{-x^{2}}<2, \forall x \in[0,1] \ & \Rightarrow \quad \int_{0}^{1}\left(1+e^{-x}\right) d x<\int_{0}^{1}\left(1+e^{-x^{2}}\right) d x<\int_{0}^{1} 2 d x \ & \Rightarrow \quad 2-\frac{1}{e}<\int_{0}^{1}\left(1+e^{-x^{2}}\right) d x<2 \end{aligned} $$
