Integral maximized 2

Calculus Level 3

Let f ⁣ : [ 0 , 1 ] → R f \colon [0, 1] \to \mathbb R be a continuous function, then find the maximum value of ∫ 0 1 x 2 f ( x ) d x − ∫ 0 1 x ( f ( x ) ) 2 d x \int_{0}^{1} x^{2}f(x) \,dx - \int_{0}^{1} x(f(x))^{2} \,dx


The answer is 0.0625.

Danish Ahmed by Danish Ahmed , 24, India

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1 solution

Nick Kent
Apr 10, 2021

The function under the integral: x 2 ∗ f − x ∗ f 2 = − x ∗ ( f 2 − 2 ∗ x 2 ∗ f ) = − x ∗ ( f 2 − 2 ∗ x 2 ∗ f + ( x 2 ) 2 ) + x 3 4 = − x ∗ ( f − x 2 ) 2 + x 3 4 ≤ x 3 4 {x}^{2}*f - x*{f}^{2} = -x * ({f}^{2} - 2 * \frac{x}{2} * f) = -x * ({f}^{2} - 2 * \frac{x}{2} * f + {(\frac{x}{2})^{2}}) + \frac{{x}^{3}}{4} = -x * {(f - \frac{x}{2})}^{2}+ \frac{{x}^{3}}{4} \leq \frac{{x}^{3}}{4} . The equality stands if f ( x ) = x 2 f(x) = \frac{x}{2} .

Then the integral itself is less or equal than ∫ 0 1 x 3 4 d x = 1 16 = 0.0625 \int_{0}^{1} \frac{{x}^{3}}{4} \,dx = \frac{1}{16} = \boxed{0.0625}

0 Helpful 0 Interesting 1 Brilliant 0 Confused

The first inequality holds since x is in [0; 1], so it is always positive

Nick Kent - 2 months ago

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