The value of int_{0}^{1} |3x^2 - 1| dx is | vedclass

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

Question

(B) We need to evaluate the integral $I = \int_{0}^{1} |3x^2 - 1| dx$. First,we find the critical point where $3x^2 - 1 = 0$,which gives $x^2 = 1/3$,s

Vedclass · Accepted Answer

$4/(3\sqrt{3})$

Vedclass · Answer

(B) We need to evaluate the integral $I = \int_{0}^{1} |3x^2 - 1| dx$. First,we find the critical point where $3x^2 - 1 = 0$,which gives $x^2 = 1/3$,so $x = 1/\sqrt{3}$ (since $x \in [0, 1]$). For $0 \le x For $1/\sqrt{3}  0$,so $|3x^2 - 1| = 3x^2 - 1$. Thus,$I = \int_{0}^{1/\sqrt{3}} (1 - 3x^2) dx + \int_{1/\sqrt{3}}^{1} (3x^2 - 1) dx$. Evaluating the integrals: $I = [x - x^3]_{0}^{1/\sqrt{3}} + [x^3 - x]_{1/\sqrt{3}}^{1}$. $I = (\frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}}) - (0) + (1 - 1) - (\frac{1}{3\sqrt{3}} - \frac{1}{\sqrt{3}})$. $I = \frac{2}{3\sqrt{3}} - (\frac{1-3}{3\sqrt{3}}) = \frac{2}{3\sqrt{3}} + \frac{2}{3\sqrt{3}} = \frac{4}{3\sqrt{3}}$.

The value of $\int_{0}^{1} |3x^2 - 1| dx$ is

Similar Questions

$\int_0^{\pi / 2} \sin ^5\left(\frac{x}{2}\right) \cdot \sin x \, dx =$

The number of continuous functions $f:[0,1] \rightarrow \mathbb{R}$ that satisfy $\int_0^1 x f(x) dx = \frac{1}{3} + \frac{1}{4} \int_0^1 (f(x))^2 dx$ is

Number of ordered pairs $(a, b)$ satisfying simultaneously the system of equations $\int_{a}^{b} x^{3} dx = 0$ and $\int_{a}^{b} x^{2} dx = \frac{2}{3}$ is

$\int_2^5 (\sqrt{x+2 \sqrt{x-1}} + \sqrt{x-2 \sqrt{x-1}}) dx = $ (in $/3$)

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module