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Calculus Level 4

∫ 0 y x 4 + ( y − y 2 ) 2 d x \begin{aligned} \int_{0}^{y} \sqrt{x^4+(y-y^2)^2}dx \end{aligned}

Find the maximum value of the above expression for 0 ≤ y ≤ 1 0 \leq y \leq 1 .

3 5 \frac{3}{5} 1 3 \frac{1}{3} 5 3 \frac{5}{3} 2 7 \frac{2}{7}

Vishruth Bharath by Vishruth Bharath , 17, USA

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1 solution

Alan Yan
Jan 27, 2018

Let f ( y ) = ∫ 0 y x 4 + ( y − y 2 ) 2 d x . f(y) = \int_{0}^{y} \sqrt{x^4+(y-y^2)^2}dx. Then, f ′ ( y ) = y 4 + y 2 ( 1 − y ) 2 f'(y) = \sqrt{y^4 + y^2(1-y)^2} . Since f ′ ( y ) ≥ 0 f'(y) \geq 0 it is monotonically increasing and maximized at y = 1 y = 1 . This gives us 1 / 3 1/3 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

This is not true. We have to differentiate inside the integral sign as well, so that f ′ ( y ) = y y 2 + ( 1 − y ) 2 + ∫ 0 y y ( 1 − y ) ( 1 − 2 y ) x 4 + ( y − y 2 ) 2 d x f'(y) \; = \; y\sqrt{y^2 + (1-y)^2} + \int_0^y \frac{y(1-y)(1-2y)}{\sqrt{x^4 + (y-y^2)^2}}\,dx and it is not obvious that this is always positive, since 1 − 2 y 1 - 2y can be negative.

Mark Hennings - 3 years, 4 months ago

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