Visual SolutionPYQ 2024 · Jan Shift 2Easy

Question

The value of

\int_0^1 (2x^3 - 3x^2 - x + 1)^{1/3} \, dx

is equal to:

(A)

0

(B)

1

(C)

2

(D)

-1

Solution Path

f(1-x) = -f(x)

means odd symmetry about

x=1/2

. King's property gives

I = -I

, so

I = 0

01Question Setup

1/4

Evaluate

\int_0^1 (2x^3 - 3x^2 - x + 1)^{1/3} \, dx

Evaluate the definite integral

02Symmetry Check

2/4

Let

f(x) = 2x^3 - 3x^2 - x + 1

. Compute

f(1-x)

by expanding - it simplifies to

-f(x)

f(1-x) = -f(x)

03King's PropertyKEY INSIGHT

3/4

By King's property:

I = \int_0^1 f(1-x)^{1/3} dx = \int_0^1 (-f(x))^{1/3} dx = -I

. So

2I = 0

I = -I \implies 2I = 0 \implies I = 0

04Final Answer

4/4

Odd symmetry about

x = 1/2

forces the integral to vanish.

\int_0^1 (2x^3 - 3x^2 - x + 1)^{1/3} dx = \boxed{0}

- Answer (A)

Concepts from this question2 concepts unlocked

★

EASY

For definite integrals, replace x with (a+b-x) to exploit symmetry

\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx

If f(a+b-x) = -f(x), the integral is zero. If f(a+b-x) = f(x), the integral doubles

Symmetry integralsTrigonometric integralsPolynomial integrals

EASY

For symmetric limits, odd functions integrate to zero and even functions double

\int_{-a}^{a} f(x)\,dx = \begin{cases} 0 & f(-x) = -f(x) \\ 2\int_0^a f(x)\,dx & f(-x) = f(x) \end{cases}

Saves computation by detecting when an integral vanishes or can be simplified

Definite integralsTrigonometric integrals

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