If {I_1} = int_0^1 {2^{x^2}} dx,{I_2} = int_0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For $0 x^3$. Since the base $2 > 1$,the function $f(x) = 2^x$ is strictly increasing,so $2^{x^2} > 2^{x^3}$ for $0 < x < 1

Vedclass · Accepted Answer

${I_1} > {I_2}$

Vedclass · Answer

(D) For $0  x^3$. Since the base $2 > 1$,the function $f(x) = 2^x$ is strictly increasing,so $2^{x^2} > 2^{x^3}$ for $0 Integrating both sides from $0$ to $1$,we get $\int_0^1 2^{x^2} dx > \int_0^1 2^{x^3} dx$,which implies ${I_1} > {I_2}$.For $1  x^2$. Similarly,$2^{x^3} > 2^{x^2}$ for $1 Integrating both sides from $1$ to $2$,we get $\int_1^2 2^{x^3} dx > \int_1^2 2^{x^2} dx$,which implies ${I_4} > {I_3}$.Thus,the correct statement is ${I_1} > {I_2}$.

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 which of the following is true?

Similar Questions

$\int_{1}^{3} \left(\frac{x^{2}+1}{4x}\right)^{-1} dx = $ . . . . . . .

The value of $b > 3$ for which $12 \int \limits_{3}^{b} \frac{1}{(x^{2}-1)(x^{2}-4)} dx = \log _{e}(\frac{49}{40})$ is equal to

If $\int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \,d x=\frac{k}{6}$, then the value of $k$ is

The value $9 \int_0^9 \left[ \sqrt{\frac{10x}{x+1}} \right] dx$,where $[t]$ denotes the greatest integer less than or equal to $t$,is . . . . . . .

The value of the integral $\int_{0}^{1} e^{x^{2}} d x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module