Mathematics · Definite Integration

JEE Main 2024 — 1 February, Shift 2 — Question 4

The value of $\int_{0}^{1}\left(2 x^{3}-3 x^{2}-x+1\right)^{\frac{1}{3}} d x$ is equal to:

Option A:
0
Correct
Option B:
1
Option C:
2
Option D:
-1

Answer: A

Step-by-step solution

$\quad I=\int_{0}^{1}\left(2 x^{3}-3 x^{2}-x+1\right)^{\frac{1}{3}} d x$ $\operatorname{Using} \int_{0}^{2 a} f(x) d x=0$ where $f(2 a-x)=-f(x)$

Here $\quad f(1-x)=-f(x)$

$\therefore \mathrm{I}=0$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Mathematics
Chapter: Definite Integration
Topic: Methods of solving definite integrals(kings rule,odd even)

The value of int 0 1 (2 x 3 -3 x 2 -x+1 ) 1/3 d x is equal to: | JEE Main 2024 PYQ with Solution · DhiX AI