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

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(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

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

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(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

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