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MCQMediumJEE 2024Properties of Definite Integrals

JEE Mathematics 2024 Question with Solution

The value of ∫01(2x3−3x2−x+1)1/3 dx\int_0^1 \left(2x^3 - 3x^2 - x + 1\right)^{1/3} \, dx is equal to:

  • A

    00

  • B

    11

  • C

    22

  • D

    −1-1

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

I=∫01(2x3−3x2−x+1)1/3 dxI = \int_0^1 \left(2x^3 - 3x^2 - x + 1\right)^{1/3} \, dx

Find: The value of II.

Using King's property of definite integrals, replace xx by 1−x1-x:

I=∫01[2(1−x)3−3(1−x)2−(1−x)+1]1/3 dxI = \int_0^1 \left[2(1-x)^3 - 3(1-x)^2 - (1-x) + 1\right]^{1/3} \, dx

Now simplify the expression inside the bracket:

2(1−x)3−3(1−x)2−(1−x)+1=−2x3+3x2+x−12(1-x)^3 - 3(1-x)^2 - (1-x) + 1 = -2x^3 + 3x^2 + x - 1

Therefore,

I=∫01(−2x3+3x2+x−1)1/3 dxI = \int_0^1 \left(-2x^3 + 3x^2 + x - 1\right)^{1/3} \, dx

Since

−2x3+3x2+x−1=−(2x3−3x2−x+1)-2x^3 + 3x^2 + x - 1 = -\left(2x^3 - 3x^2 - x + 1\right)

we get

I=−∫01(2x3−3x2−x+1)1/3 dx=−II = -\int_0^1 \left(2x^3 - 3x^2 - x + 1\right)^{1/3} \, dx = -I

Hence,

2I=0⇒I=02I = 0 \Rightarrow I = 0

Therefore, the value of the integral is 00. The correct option is A.

Expanded Simplification

Given:

I=∫01(2x3−3x2−x+1)1/3 dxI = \int_0^1 \left(2x^3 - 3x^2 - x + 1\right)^{1/3} \, dx

Find: Evaluate the definite integral.

Apply the substitution used in King's rule:

I=∫01[2(1−x)3−3(1−x)2−(1−x)+1]1/3 dxI = \int_0^1 \left[2(1-x)^3 - 3(1-x)^2 - (1-x) + 1\right]^{1/3} \, dx

Expand each term:

(1−x)3=1−3x+3x2−x3(1-x)^3 = 1 - 3x + 3x^2 - x^3 (1−x)2=1−2x+x2(1-x)^2 = 1 - 2x + x^2

So,

2(1−x)3=2−6x+6x2−2x32(1-x)^3 = 2 - 6x + 6x^2 - 2x^3 −3(1−x)2=−3+6x−3x2-3(1-x)^2 = -3 + 6x - 3x^2 −(1−x)=−1+x-(1-x) = -1 + x

Combining all terms,

2−6x+6x2−2x3−3+6x−3x2−1+x+1=−1+x+3x2−2x32 - 6x + 6x^2 - 2x^3 - 3 + 6x - 3x^2 - 1 + x + 1 = -1 + x + 3x^2 - 2x^3

Thus,

I=∫01(−1+x+3x2−2x3)1/3 dxI = \int_0^1 \left(-1 + x + 3x^2 - 2x^3\right)^{1/3} \, dx

Rewrite the polynomial as

−1+x+3x2−2x3=−(2x3−3x2−x+1)-1 + x + 3x^2 - 2x^3 = -\left(2x^3 - 3x^2 - x + 1\right)

Therefore,

I=−∫01(2x3−3x2−x+1)1/3 dx=−II = -\int_0^1 \left(2x^3 - 3x^2 - x + 1\right)^{1/3} \, dx = -I

Hence,

2I=0⇒I=02I = 0 \Rightarrow I = 0

Therefore, the final answer is 00.

Common mistakes

  • Students may apply the property ∫01f(x) dx=∫01f(1−x) dx\int_0^1 f(x) \, dx = \int_0^1 f(1-x) \, dx but forget to simplify the transformed polynomial completely. This hides the fact that the new integrand is the negative of the original one. Always expand and collect like terms carefully.

  • A common mistake is to write (−a)1/3=(a)1/3\left(-a\right)^{1/3} = \left(a\right)^{1/3}. This is wrong because cube root preserves the sign: (−a)1/3=−a1/3\left(-a\right)^{1/3} = -a^{1/3}. Use this sign property correctly to obtain I=−II = -I.

  • Some students confuse King's rule with symmetry formulas meant for f(x)+f(1−x)f(x)+f(1-x). Here the key step is comparing the transformed integrand with the original integrand, not adding two unrelated expressions. First rewrite the integrand after replacing xx by 1−x1-x, then identify the sign change.

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