The value of $\int\limits_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} \mathrm{~d} x$ is equal to :
Solution
$\begin{aligned} & I=\int_0^1\left(2 x^3-3 x^2-x+1\right)^{1 / 3} d x \\\\ & I=\int_0^1\left((2 x-1)\left(x^2-x-1\right)\right)^{1 / 3} d x \\\\ & I=\int_0^1\left[(2(1-x)-1)\left((1-x)^2-(1-x)-1\right)\right]^{1 / 3} d x \\\\ & I=\int_0^1\left((1-2 x)\left(x^2-x-1\right)\right)^{1 / 3} d x\end{aligned}$
<br/><br/>$\begin{aligned} & I=-\int_0^1\left((2 x-1)\left(x^2-x-1\right)\right)^{1 / 3} d x \\\\ & I=-I \\\\ & 2 I=0 \\\\ & I=0\end{aligned}$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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