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If $\displaystyle f(x) = x^3+\frac{3x}{4}-\frac{3x^2}{2}+\frac{7}{8}.$ Then $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx$

Try: Given $\displaystyle f(x) =x^3+\frac{3x}{4}-\frac{3x^2}{2}+\frac{7}{8}. $

Then $\displaystyle f(1-x) = (1-x)^3+\frac{3}{4}(1-x)-\frac{3}{2}(1-x)^2+\frac{7}{8}$

$$f(1-x) = -x^3+\frac{3x^2}{2}-\frac{3x}{4}+\frac{9}{8}$$

$$f(x)+f(1-x) = 2$$

Let $$I = \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx = \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(1-x))dx$$

I did not understand how to solve from there

could some help me to solve it

DXT
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  • I mean, is there much to be said about this? Just compute $f(f(x))$ and then compute the integral. – MathematicsStudent1122 Jan 12 '19 at 07:43
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    Notice $$f(x) = \left(x-\frac12\right)^3 + 1\implies f(f(x)) = \left(\left(x-\frac12\right)^3 + \frac12\right)^3 + 1$$ Change variable to $u = x - \frac12$, the integral becomes

    $$\int_{-\frac14}^\frac14 \left[\left(u^3 + \frac12\right)^3 + 1\right] du \stackrel{\color{red}{\text{WHY?}}}{=} 2\int_0^\frac14 \left[\frac32 u^6 + \frac98\right] du = \cdots $$

     – achille hui Jan 12 '19 at 07:53

1 Answers1

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Hint

Note that $$f(x)={(x-{1\over 2})^3}+1$$and $$\int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx=\int^{\frac{1}{4}}_{-\frac{1}{4}}f(f(u+{1\over 2}))du=\int^{\frac{1}{4}}_{-\frac{1}{4}}f(u^3+1)du$$

Mostafa Ayaz
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