Question

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

Answer 1

First we have to define what the terms we need to solve the problem are,
Since the values of the integral are getting changed in different integrals like from zero to one can be changed as the two to one.
So first we have to find which integrals are greater than which integral so that we can solve this problem so easily.

Complete step by step answer:
Let there are four given integrals ${I_1},{I_2},{I_3},{I_4}$ with different values and power two and three.
Since as we know that $x$ is greater than ${x^2}$ and also ${x^2}$ is greater than ${x^3}$ .
Thus, by using the help of these values we can simply eliminate the integral ${I^3},{I^2}$ (the x power two is the lowest among all). Hence now we have left with three similar integrals in power but not in integral values.
Since integral over two to one is always less than the integral over one to zero (because acting on the zero does not affect too much on the values); $(0,1) > (1,2)$ so the integral over one to zero is the greatest integral among all.
Apply for a simple integral like $\int\limits_0^1 x dx = [\dfrac{{{x^2}}}{2}]_0^1 = \dfrac{1}{2}$ and $\int\limits_1^2 x dx = [\dfrac{{{x^2}}}{2}]_1^2 = \dfrac{4}{2} - \dfrac{1}{2} = \dfrac{3}{2}$
(But for ${2^{{x^3}}}$ is greater than ${2^{{x^2}}}$ in $(0,1) > (1,2)$ )
Thus, we get ${I_1},{I_2}$ are possible but the integral two is already eliminated as the power value is low.
Hence, we get the integral one is the greatest among all and thus option $A){I_1} > {I_2}$ is correct (because integral two is greater than three)

So, the correct answer is “Option A”.

Note: we are also able to solve this problem by just applying the integrals formula over one to zero and over two to one.
And the general form of the integrals can be framed as ${I_1} > {I_2} > {I_3} > {I_4}$ integral one is the greatest among them all because it satisfies both the power terms higher and greater integrals values among all.