Question

If \[{I_1} = \int\limits_0^1 {{2^{{x^2}}}dx} ,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx} ,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} \] and \[{I_4} = \int\limits_1^2 {{2^{{x^3}}}dx} \] then
(A) \[{I_3} > {I_4}\]
(B) \[{I_3} = {I_4}\]
(C) \[{I_1} > {I_2}\]

Answer 1

Hint:
Firstly, we will check that for the integrals \[{I_1}\] and \[{I_2}\], which function gives larger value in the interval \[[0,1]\]. Then we will check the same for the integrals \[{I_3}\] and \[{I_4}\]. And on comparing, we will get the result.

Complete step by step solution:
First, we will take the interval \[{I_1}\] and \[{I_2}\] because they have the same interval 0 to 1.
Square of any value from the interval 0 to 1 is greater than the cube of that value.
 \[ \Rightarrow {x^2} > {x^3}\] in the interval \[[0,1]\].
Also, we can see that
i.e. \[\int {{x^2}} > \int {{x^3}} \] in the interval \[[0,1]\].
We can also see that
 \[ \Rightarrow {2^{{x^2}}} > {2^{{x^3}}}\] in the interval \[[0,1]\].
Therefore, the value of \[\int\limits_0^1 {{2^{{x^2}}}dx} \] is greater than the value of \[\int\limits_0^1 {{2^{{x^3}}}} dx\].
 \[ \Rightarrow \int\limits_0^1 {{2^{{x^2}}}dx} > \int\limits_0^1 {{2^{{x^3}}}} dx\]
Hence, \[{I_1} > {I_2}\].
Now, we can see that the cube of any value in the interval \[[1,2]\] is greater than the square of that value.
The value of \[{x^3}\] in the interval \[[1,2]\] is larger than the value of \[{x^2}\].
Hence, \[{x^3} > {x^2}\] in the interval \[[1,2]\].
The integration of \[{x^3}\] is greater than the integration of \[{x^2}\] in the interval \[[1,2]\].
Hence,
 \[ \Rightarrow \int {{x^3}} > \int {{x^2}} \] in the interval \[[1,2]\].
Therefore, the value of \[\int\limits_1^2 {{2^{{x^3}}}dx} \] is greater than the value of \[\int\limits_1^2 {{2^{{x^2}}}} dx\].
 \[ \Rightarrow \int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}} dx\]
Hence, \[{I_4} > {I_3}\].
Therefore \[{I_3} \ne {I_4}\]

Hence option (C) is correct
i.e. \[{I_1} > {I_2}\]

Note:
The given integrals are definite integrals. Definite integrals are the integrals whose limits are given. The definite integrals are mainly used to calculate the area of the region.