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

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Definite Integrals

The value of

\int_{0}^{1} (2 x^{3} - 3 x^{2} - x + 1)^{\frac{1}{3}} d x

is equal to:

Answer: (A)

Let

g (x) = 2 x^{3} - 3 x^{2} - x + 1

, so the integrand is

f (x) = g (x)^{1/3}

.

Check

g (1 - x)

: substituting and expanding,

g (1 - x) = 2 (1 - x)^{3} - 3 (1 - x)^{2} - (1 - x) + 1 = - 2 x^{3} + 3 x^{2} + x - 1 = - g (x) .

Since the cube root is an odd function,

f (1 - x) = (g (1 - x))^{1/3} = (- g (x))^{1/3} = - f (x)

.

So

f (x) + f (1 - x) = 0

for every

x \in [0, 1]

.

Now, using the substitution

x \to 1 - x

in the definite integral,

I = \int_{0}^{1} f (x) d x = \int_{0}^{1} f (1 - x) d x .

Adding the two expressions for

I

2 I = \int_{0}^{1} [f (x) + f (1 - x)] d x = \int_{0}^{1} 0 d x = 0.

I = 0.

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