The value of ∫(0 to 1) (2x³ - 3x² - x + 1)^(1/3) dx is equal to:
(1) 0
(2) 1
(3) 2
(4) -1

Q: Why can't I just use substitution or integration by parts to solve this?

Direct integration is a trap! The cubic polynomial inside the cube root is designed to be impossible to integrate using standard techniques. The problem is a test of your ability to recognize symmetry, not your ability to perform brute-force calculus. Always look for the 'King's Property' when you see limits from $0$ to $a$.

Q: How do I know when to use the King's Property?

Whenever you see a definite integral with limits from $0$ to $a$, especially when the integrand looks like a nightmare, the King's Property is your first line of defense. It transforms the variable of integration without changing the value of the integral, often revealing a hidden symmetry that makes the problem collapse instantly.

Q: Wait, why does $I = -I$ imply $I=0$?

It is a simple algebraic consequence! If $I = -I$, then adding $I$ to both sides gives $2I = 0$. Since $2$ is not zero, $I$ must be zero. It is the mathematical equivalent of saying 'the positive area and negative area perfectly cancel out'.

Question

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

Options:

(A) 0

(B) 1

(C) 2

(D) -1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to evaluate the definite integral:

$$I = \int_{0}^{1} (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} dx$$

At first glance, the cubic polynomial trapped inside a cube root appears daunting. However, in JEE Advanced mathematics, such complexity often masks an elegant underlying symmetry.

### The Secret Key: The King's Property

We utilize the 'King's Property', a fundamental tool in definite integration. It states that for any function $f(x)$ integrated over the interval $[0, a]$:

$$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$$

This property allows us to replace $x$ with $(a-x)$ without altering the value of the integral. In our specific case, $a=1$, so we will substitute $x$ with $(1-x)$.

### The Algebraic Grind

Let our integrand be $f(x) = (2x^3 - 3x^2 - x + 1)^{\frac{1}{3}}$. Applying the substitution $x 	o (1-x)$, we obtain:

$$f(1-x) = [2(1-x)^3 - 3(1-x)^2 - (1-x) + 1]^{\frac{1}{3}}$$

Expanding the binomial terms $(1-x)^3 = 1 - 3x + 3x^2 - x^3$ and $(1-x)^2 = 1 - 2x + x^2$, we substitute these into the expression:

$$[2(1 - 3x + 3x^2 - x^3) - 3(1 - 2x + x^2) - (1-x) + 1]^{\frac{1}{3}}$$

Distributing the constants and grouping the terms by powers of $x$:

$$[ (2 - 6x + 6x^2 - 2x^3) + (-3 + 6x - 3x^2) - 1 + x + 1 ]^{\frac{1}{3}}$$

Combining the coefficients for each power of $x$:

$$(-2x^3 + 3x^2 + x - 1)^{\frac{1}{3}}$$

### The Revelation

Factoring out a negative sign from the expression, we observe:

$$f(1-x) = -(2x^3 - 3x^2 - x + 1)^{\frac{1}{3}} = -f(x)$$

This reveals that the function possesses odd symmetry about the midpoint $x = 0.5$. Consequently, the area under the curve from $0$ to $0.5$ is exactly the negative of the area from $0.5$ to $1$.

Applying the King's Property to our integral $I$:

$$I = \int_{0}^{1} f(1-x) dx = \int_{0}^{1} -f(x) dx = -I$$

Since $I = -I$, it follows that $2I = 0$. Therefore, the final value of the integral is:

**$I = 0$**

Animated Solution for Mathematics - Definite Integration: The value of $\int_{0}^{1} (2 x^{3} - 3 x^{2} - x + 1)^{\frac{1}{3}} d x$ is equal to:

Select Answer:

Visualized Solution

Defining the Integral $I$

The King's Property

Applying the Property

Expanding the Binomials

Substituting the Expansions

Distributing the Constants

Grouping Like Terms

Simplifying $f (1 - x)$

Relating Back to $f (x)$

Visualizing the Symmetry

Evaluating the Integral

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Secret Key

The Algebraic Grind

The Revelation

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ∫01​(2x3−3x2−x+1)31​dx is equal to: