Question

The value of the integral $\int\limits_0^1 {{e^{{x^2}}}} dx$ is-
A) Less than e
B) Greater than e
C) Less than $1$
D) Greater than $1$

Answer 1

We can observe from the function that$0 < x < 1$. Now if we choose any arbitrary number between this interval we will find that the square of the number will be less than the number itself. So we can write$1 > x > {x^2}$. Now we know that, ${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + ... + \dfrac{{{x^n}}}{{n!}}$ so if $x > {x^2}$ then the exponential of x will also be greater than that of ${x^2}$. Then integrate both sides and solve to get an inequality. Again we can write, $1 < {e^{{x^2}}}$because we know that $1 < {e^x}$. Then follow the same process to get an inequality. Join the inequalities together to get the answer.

Complete step by step solution:
We have to find the value of$\int\limits_0^1 {{e^{{x^2}}}} dx$.
Let I=$\int\limits_0^1 {{e^{{x^2}}}} dx$
So the lower limit is $0$ and upper limit is $1$. Then x lies between the upper limit and lower limit, so we can write-
$\Rightarrow 0 < x < 1$ -- (i)
Now suppose we choose any arbitrary number between these two limits: suppose x=$0.5$ then its square will be-${x^2} = 0.25$. So if we choose any number between these limits, the square of the number will be less than the number itself.
Then we can say that $x > {x^2}$ -- (ii)
Then from eq. (i) and (ii), x is greater than ${x^2}$ and it is also smaller than $1$so we get-
$\Rightarrow 1 > x > {x^2}$ -- (iii)
Then the exponential of x will also be greater than exponential of ${x^2}$ so, we can also write-
$\Rightarrow {e^{{x^2}}} < {e^x}$ {From eq. (ii)}
On integrating both sides with upper limit $0$ and lower limit$1$, we get-
$\Rightarrow \int\limits_0^1 {{e^{{x^2}}}} dx < \int\limits_0^1 {{e^x}} dx$
Now we know that $\int\limits_a^b {{e^x}} dx = \left[ {{e^x}} \right]_a^b$ so we get-
$\Rightarrow {\text{I}} < \left[ {{e^x}} \right]_0^1$
On putting upper limit and lower limit in place of x, we get-
$\Rightarrow {\text{I}} < \left[ {{e^1} - {e^0}} \right]$
Now, we know that ${e^0} = 1$ so we get-
$\Rightarrow {\text{I}} < \left[ {e - 1} \right]$-- (iv)
Now we know that $1 < {e^x}$ as ${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + ... + \dfrac{{{x^n}}}{{n!}}$
Then we can write,
$\Rightarrow 1 < {e^{{x^2}}}$ {as ${e^{{x^2}}} < {e^x}$ }
On integrating both sides, we get-
$\Rightarrow \int\limits_0^1 {1dx} < \int\limits_0^1 {{e^{{x^2}}}dx}$
On solving, we get-
$\Rightarrow \left[ x \right]_0^1 < I$
On putting lower limit and upper limit, we get-
$\Rightarrow 1 - 0 < I$
Then we get-
$\Rightarrow 1 < I$ -- (v)
From eq. (iv) and (v), we get-$1 < I < e - 1$
The value of the integral is greater than $1$ and less than$e$.

Answer-The correct options are option A and option D.

Note:
Here if the student tries to integrate the function directly then the function obtained will be too complex to get the answer easily-
First we will assume ${x^2} = t$
On differentiation, we get-
$\Rightarrow 2xdx = dt$
$\Rightarrow dx = \dfrac{{dt}}{{2\sqrt t }}$
On putting this value in the function, we get-
$\Rightarrow \dfrac{1}{2}\int\limits_0^1 {{e^t}} \dfrac{1}{{\sqrt t }}dt$
Now we will have to use the product rule which will again give us a product of two functions because $\int {{e^x}} dx = {e^x}$ . This will take too long to solve and will make the calculation complex.
Hence we have solved the question in terms of inequality.