JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of \(\int_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} d x\) is equal to :
- A \(0\)
- B \(1\)
- C \(2\)
- D \(-1\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\(I=\int_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} d x\) Using \(\int_0^{2 a} f(x) d x=0\) where \(f(2 a-x)=-f(x)\) Here \(\quad f(1-x)=-f(x)\) \(\therefore \mathrm{I}=0\)
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