Question

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

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

Answer 1

The correct option is A $I_1>I_2$

$I_1={\int }_0^1{2}^{x^2}dx;I_2={\int }_0^1{2}^{x^3}dx;I_3={\int }_1^2{2}^{x^2}dx$
$I_4={\int }_1^2{2}^{x^2}dx$
If ${\int }_a^{b}f\left(x\right)>{\int }_a^{b}g\left(x\right)$ then we know that $f\left(x\right)>g\left(x\right)$ in $[a,b]$
So, $2^{x^3}[1,2]>2^{x^2}[1,2]>2^{x^2}[0,1]>2^{x^3}[0,1]$
$I_4>I_3>I_1>I_2$
So, checking the relation
$I_1>I_2$ is only correct.