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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\...

If I1=012x2dx,I2=012x3dx,I3=122x2dx{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 I4=122x3dx{I_4} = \int\limits_1^2 {{2^{{x^3}}}dx} then
(A) I3>I4{I_3} > {I_4}
(B) I3=I4{I_3} = {I_4}
(C) I1>I2{I_1} > {I_2}

Explanation

Solution

Firstly, we will check that for the integrals I1{I_1} and I2{I_2}, which function gives larger value in the interval [0,1][0,1]. Then we will check the same for the integrals I3{I_3} and I4{I_4}. And on comparing, we will get the result.

Complete step by step solution:
First, we will take the interval I1{I_1} and I2{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.
x2>x3\Rightarrow {x^2} > {x^3} in the interval [0,1][0,1].
Also, we can see that
i.e. x2>x3\int {{x^2}} > \int {{x^3}} in the interval [0,1][0,1].
We can also see that
2x2>2x3\Rightarrow {2^{{x^2}}} > {2^{{x^3}}} in the interval [0,1][0,1].
Therefore, the value of 012x2dx\int\limits_0^1 {{2^{{x^2}}}dx} is greater than the value of 012x3dx\int\limits_0^1 {{2^{{x^3}}}} dx.
012x2dx>012x3dx\Rightarrow \int\limits_0^1 {{2^{{x^2}}}dx} > \int\limits_0^1 {{2^{{x^3}}}} dx
Hence, I1>I2{I_1} > {I_2}.
Now, we can see that the cube of any value in the interval [1,2][1,2] is greater than the square of that value.
The value of x3{x^3} in the interval [1,2][1,2] is larger than the value of x2{x^2}.
Hence, x3>x2{x^3} > {x^2} in the interval [1,2][1,2].
The integration of x3{x^3} is greater than the integration of x2{x^2} in the interval [1,2][1,2].
Hence,
x3>x2\Rightarrow \int {{x^3}} > \int {{x^2}} in the interval [1,2][1,2].
Therefore, the value of 122x3dx\int\limits_1^2 {{2^{{x^3}}}dx} is greater than the value of 122x2dx\int\limits_1^2 {{2^{{x^2}}}} dx.
122x3dx>122x2dx\Rightarrow \int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}} dx
Hence, I4>I3{I_4} > {I_3}.
Therefore I3I4{I_3} \ne {I_4}

**Hence option (C) is correct
i.e. I1>I2{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.