Question

If $I_1={\int }_0^1{2}^{x^2}dx,I_2={\int }_0^1{2}^{x^3}dx,I_3={\int }_1^2{2}^{x^2}dx$ and $I_4={\int }_1^2{2}^{x^3}dx$, then

• A. $I_3>I_4$
• B. $I_3=I_4$
• C. $I_1>I_2$
• D. $I_2>I_1$

Answer 1

The correct option is C $I_1>I_2$

${\int }_0^1{2}^{x^2}dx,{\int }_0^1{2}^{x^3}dx,{\int }_0^1{2}^{x^3}dx,{\int }_0^1{2}^{x^2}dx$

On solving ${\int }_0^1{2}^{x^3},$ we get the answer as

$\frac{-1}{3}{E}_{2/3}(-log(2\left)\right)-\frac{(-1{)}^{2/3}\sqrt{\left(4/3\right)}}{\sqrt[3]{3log\left(2\right)}}$

$I_2\approx 1,21399$

& ${\int }_0^1{2}^{x^3}dx=\frac{1}{3}\left[E_{2/3}\right(-log\left(2\right))-2E_{2/3}(-8log\left(2\right)\left)\right]$

$I_4\approx 35.9324$

& ${\int }_0^1{2}^{x^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{log\left(2\right)}}\left(eyi\right(2\sqrt{log\left(2\right)}-eyi\left(log2\right)\left)\right)$

$I_3\approx 6.05225$

& ${\int }_0^1{2}^{x^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{log\left(2\right)}}eyi\left(log2\right)\approx 1,28823=I_1$

from the values of $I_1,I_2,I_3$ & $I_4,$ we can

conclude that $I_1>I_2$