int_0^{1.5} x[x^2] dx = | vedclass

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

Question

(C) We need to evaluate the integral $I = \int_0^{1.5} x[x^2] dx$. Since $[x^2]$ changes its value at $x^2 = 1$ and $x^2 = 2$,we split the integral at

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int_0^{1.5} x[x^2] dx$. Since $[x^2]$ changes its value at $x^2 = 1$ and $x^2 = 2$,we split the integral at $x = 1$ and $x = \sqrt{2}$. $I = \int_0^1 x[x^2] dx + \int_1^{\sqrt{2}} x[x^2] dx + \int_{\sqrt{2}}^{1.5} x[x^2] dx$. For $0 \le x For $1 \le x For $\sqrt{2} \le x Thus,$I = \int_0^1 x(0) dx + \int_1^{\sqrt{2}} x(1) dx + \int_{\sqrt{2}}^{1.5} x(2) dx$. $I = 0 + \left[ \frac{x^2}{2} \right]_1^{\sqrt{2}} + 2 \left[ \frac{x^2}{2} \right]_{\sqrt{2}}^{1.5}$. $I = \left( \frac{2-1}{2} \right) + (1.5^2 - (\sqrt{2})^2) = \frac{1}{2} + (2.25 - 2) = 0.5 + 0.25 = 0.75 = \frac{3}{4}$.

$\int_0^{1.5} x[x^2] dx = $

Similar Questions

$\int_{-1}^{2} {|x|\,dx} =$ (in $/2$)

If $[2,6]$ is divided into four intervals of equal length,then the approximate value of $\int_2^6 \frac{1}{x^2-x} dx$ using Simpson's rule is

$\int_0^2 \sqrt{(x+3)(2-x)} \, dx =$

$\int_0^{\pi /4} {\frac{{4\sin 2\theta \,d\theta }}{{{{\sin }^4}\theta + {{\cos }^4}\theta }}} = $

$\int_{-2}^3 |1-x^2| dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module