No one was able to solve it previously

Calculus Level 4

If f ( x ) = x 3 − 3 2 x 2 + x + 1 4 f(x)=x^{3}-\frac{3}{2}x^{2}+x+\frac{1}{4}

Find ∫ 1 4 3 4 f ( f ( x ) ) d x \displaystyle {\int^{\frac{3}{4}}_{\frac{1}{4}} f(f(x)) \text{ d}x}

This problem is part of my set Some JEE problems


The answer is 0.25.

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Tanishq Varshney by Tanishq Varshney , 23, India

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3 solutions

Tanishq Varshney
Feb 13, 2015

plz do upvote

10 Helpful 0 Interesting 1 Brilliant 0 Confused

Now there's a clever shortcut that I haven't seen before. Bravo!

Edwin Hughes - 6 years, 3 months ago

Guessed that you may have special way but I think I couldn't have found the way you described after reading your solution.

Lu Chee Ket - 6 years, 3 months ago

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i have used ∫ a b f ( x ) . d x = ∫ a b f ( a + b − x ) . d x \int^{b}_{a}f(x).dx=\int^{b}_{a}f(a+b-x).dx

Tanishq Varshney - 6 years, 3 months ago

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I have seen this in Brilliant for many times. Not obvious why negative sign is not needed at least for interchange of a and b. f(1-x) = 1 - f(x) seem to be very specific for this question and usually cannot be noticed unless getting to it.

Can you show how f(1- f(x)) = 1 - f(f(x)) and etc step by step? Could this method bring about incorrectness generally? When should we do in this way and what are the precautions?

Lu Chee Ket - 6 years, 3 months ago

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@Lu Chee Ket – first find f(1-x) which will surely come out to be 1- f(x) all ur queries will be over .Its not a property, we have just find a condition.

Tanishq Varshney - 6 years, 3 months ago

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@Tanishq Varshney – Not answered.

Lu Chee Ket - 6 years, 3 months ago
Pradeep Maurya
May 15, 2015

f ( x ) = ( x − 1 2 ) 3 + x 4 + 3 8 f(x) = \left(x-\frac{1}{2}\right)^3+\frac{x}{4}+\frac{3}{8} and ∫ 1 / 4 3 / 4 f ( f ( x ) ) d x = ∫ − 1 / 4 1 / 4 f ( f ( x + 1 2 ) ) \int_{1/4}^{3/4}f(f(x))dx = \int_{-1/4}^{1/4}f\left(f\left(x+\frac{1}{2}\right)\right)

Now f ( x + 1 2 ) = x 3 + 1 4 ( x + 1 2 ) + 3 8 = x 3 + x 4 + 1 2 f\left(x+\frac{1}{2}\right) = x^3+\frac{1}{4}\left(x+\frac{1}{2}\right)+\frac{3}{8} = x^3+\frac{x}{4}+\frac{1}{2} i.e f ( f ( x + 1 2 ) ) = ( x 3 + x 4 ) 3 + 1 4 ( x 3 + x 4 ) + 1 2 = odd function + 1 2 f\left(f\left(x+\frac{1}{2}\right)\right) = \left(x^3+\frac{x}{4}\right)^3+\frac{1}{4}\left(x^3+\frac{x}{4}\right)+\frac{1}{2} = \text{odd function}+\frac{1}{2}

So the integral equals to ∫ − 1 / 4 1 / 4 1 2 d x = 0.25 \int_{-1/4}^{1/4} \frac{1}{2}dx = \boxed{0.25}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice......

Sagar Saxenae - 4 years, 7 months ago

@Pradeep Maurya You haven't put dx on 2nd integral of 2nd line

Sahar Bano - 5 months, 2 weeks ago
Lu Chee Ket
Feb 14, 2015

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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