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

Hint: 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.