Verified Solution • 2025–26 Session moderate

QJEEM24Ma4797: JEE Main 2024 Mathematics - Class 12 JEEM24Ma4797 Solution

The value of ∫01(2x3−3x2−x+1)13 dx\int\limits_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} \mathrm{~d} x is equal to :

Multiple Choice Options

  • A -1
  • B 2
  • C 0
  • D 1

Step-by-Step Verified Solution

amp;I=∫01(2x3−3x2−x+1)1/3dxamp;I=∫01((2x−1)(x2−x−1))1/3dxamp;I=∫01[(2(1−x)−1)((1−x)2−(1−x)−1)]1/3dxamp;I=∫01((1−2x)(x2−x−1))1/3dx\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}

amp;I=−∫01((2x−1)(x2−x−1))1/3dxamp;I=−Iamp;2I=0amp;I=0\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}

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