Question

If the value of ${{I}_{1}}=\int\limits_{0}^{1}{{{2}^{{{x}^{2}}}}dx}$ and ${{I}_{2}}=\int\limits_{0}^{1}{{{2}^{{{x}^{3}}}}dx}$ and ${{I}_{3}}=\int\limits_{1}^{2}{{{2}^{{{x}^{2}}}}dx}$ and ${{I}_{4}}=\int\limits_{1}^{2}{{{2}^{{{x}^{3}}}}dx}$ then,
$\left( a \right){{I}_{3}}>{{I}_{4}}$
$\left( b \right){{I}_{3}}={{I}_{4}}$
$\left( c \right){{I}_{1}}>{{I}_{2}}$
$\left( d \right){{I}_{2}}>{{I}_{1}}$

Answer 1

To solve this question, we will first consider the integral ${{I}_{1}}$ and ${{I}_{2}}$ and observe that the limit of the integral is from 0 to 1, i.e. $x\in \left[ 0,1 \right)\Rightarrow 0 **Complete step-by-step answer:** Let us first find the relation between \[{{I}_{1}}$ and ${{I}_{2}}.$
${{I}_{1}}=\int\limits_{0}^{1}{{{2}^{{{x}^{2}}}}dx}$
${{I}_{2}}=\int\limits_{0}^{1}{{{2}^{{{x}^{3}}}}dx}$
Here, as integral varies from 0 to 1, x varies from 0 to 1.
$\Rightarrow 0\le x\le 1$
Now, for $0\le x\le 1$ we have ${{x}^{2}}>{{x}^{3}}.$
Example if x = 0.5, then ${{x}^{2}}={{\left( 0.5 \right)}^{2}}=0.25$ and ${{x}^{3}}={{\left( 0.5 \right)}^{3}}=0.125$
So, ${{x}^{2}}>{{x}^{3}}.$
Now, as ${{x}^{2}}>{{x}^{3}}$
$\Rightarrow {{2}^{{{x}^{2}}}}>{{2}^{{{x}^{3}}}}$
Applying the integral on both the sides, we have,
$\int\limits_{0}^{1}{{{2}^{{{x}^{2}}}}dx}=\int\limits_{0}^{1}{{{2}^{{{x}^{3}}}}dx}$
$\Rightarrow {{I}_{1}}>{{I}_{2}}$
So, we have the option (c) is the correct answer.
Hence, the relation is determined between ${{I}_{1}}$ and ${{I}_{2}}.$
Now, consider ${{I}_{3}}$ and ${{I}_{4}}.$
${{I}_{3}}=\int\limits_{1}^{2}{{{2}^{{{x}^{2}}}}dx}$
${{I}_{4}}=\int\limits_{1}^{2}{{{2}^{{{x}^{3}}}}dx}$
Now, here integral value is between 1 and 2. So, x varies between 1 and 2.
$\Rightarrow 1\le x\le 2$
And for this value, ${{x}^{2}}<{{x}^{3}}$ as ${{\left( 1.5 \right)}^{2}}<{{\left( 1.5 \right)}^{3}}.$
$\Rightarrow {{x}^{2}}<{{x}^{3}}$
$\Rightarrow {{2}^{{{x}^{2}}}}<{{2}^{{{x}^{3}}}}$
Applying integral on both the sides, we get,
$\Rightarrow \int\limits_{1}^{2}{{{2}^{{{x}^{2}}}}dx}=\int\limits_{1}^{2}{{{2}^{{{x}^{3}}}}dx}$
$\Rightarrow {{I}_{3}}<{{I}_{4}}$
Hence, option (a) is wrong.

So, the correct answer is “Option c”.

Note: A point to note here in this question is now, ${{x}_{1}}<{{x}_{2}}.$
$\Rightarrow {{2}^{{{x}_{1}}}}<{{2}^{{{x}_{2}}}}$
Consider the function ${{e}^{x}}$ now as $e\cong 2.57.$ e and 2 behave similarly. Consider the graph of ${{e}^{x}}.$

Then we see that for x > 0, if ${{x}_{1}}>{{x}_{2}},\Rightarrow {{e}^{{{x}_{1}}}}>{{e}^{{{x}_{2}}}}.$ Now, as e and 2 base behaves the same, we can say,
$\Rightarrow {{x}_{1}}>{{x}_{2}}$
$\Rightarrow {{2}^{{{x}_{1}}}}>{{2}^{{{x}_{2}}}}$
Hence, the confusion is cleared.