The value of ∫₀¹(2 x³-3 x²-x+1)^(1/3) d x is equal to :

Solution

Maths sir · 4 steps

Replace x by 1 − x: the cubic flips its sign, so the integral equals its own negative.

1. Use the property. For any integral from 0 to a, ∫₀ᵃ g(x) dx = ∫₀ᵃ g(a-x) dx. Here a = 1, so check what happens to the cubic f(x) = 2x³-3x²-x+1 when x becomes 1-x.
2. Expand the cubic. 2(1-x)³ = 2-6x+6x²-2x³ and -3(1-x)² = -3+6x-3x². Adding -(1-x)+1 = x gives f(1-x) = -2x³+3x²+x-1.

   f(1-x) = -(2x³-3x²-x+1) = -f(x)

3. Cube root is odd. Since ∛-t = -∛t, the integrand also flips: (f(1-x))^(1/3) = -(f(x))^(1/3). So I = ∫₀¹ (f(1-x))^(1/3)dx = -I.
4. So. I = -I gives 2I = 0, so I = 0. The value of the integral is 0, option (c).

The idea to keep. When an integrand looks impossible to integrate, try x → a - x first. If it turns into its own negative, the integral from 0 to a is zero.

Answer from the NTA final answer key. Solution by PrepWiser.