2
$\begingroup$

If $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$

$\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 \mathrm dx$

could some help me with this, thanks

  • 1
    Is there any reason to not simply expand it out? Or are you looking for something more? If expanding is fine, then wolfram gave an answer of $\frac{1}{4}$. – 2017-01-09
  • 2
    The substitution $y=x-\frac{1}{2}$ simplifies $f$ to a depressed cubic, and the integration interval to a symmetric one around $0\,$. – 2017-01-09

3 Answers 3

7

As @dxiv says in the comments, we can write $$f (x) = (x-\frac {1}{2})^3 +\frac {1}{4}x +\frac {3}{8} $$ Thus, substituting $u=x- \frac {1}{2}$ gives us $f (u)=u^3+\frac {1}{4}(u+\frac {1}{2}) +\frac {3}{8} = u^3+\frac {1}{4}u +\frac {1}{2} $. Thus, $$f (f (u)) =(u^3+\frac {1}{4}u +\frac {1}{2})^3 +\frac {1}{4}[u^3 +\frac {1}{4}u +\frac {1}{2}] +\frac {1}{2}$$ As $x $ goes from $\frac {1}{4} $ to $\frac {3}{4} $,. $u $ goes from $-\frac {1}{4} $ to $\frac {1}{4} $.

Hope you can take it from here.

2

Apply kings rule, I=int(f(f(1-x))) =integral( -x3 +1.5x2 -x +0.75)
Add the two 2I = int(1)
I= 0.25.

  • 0
    Sorry for the unformatted answer, on mobile, in journey – 2017-01-09
  • 0
    How could $\int f(f(1-x))dx=\int -x^3+1.5x^2-x+0.75 dx$ I mean $\int f(1-x)=\int -x^3+1.5x^2-x+0.75 dx$ and not the one you stated.Thanks. – 2017-04-07
1

You can use the property given here: $$\color{blue}{\int_a^b f(x)dx=\int_a^{(a+b)/2}[f(x)+f(a+b-x)]dx \Rightarrow \\ \int_a^b f(f(x))dx=\int_a^{(a+b)/2}[f(f(x))+f(f(a+b-x))]dx};\\ \int_{1/4}^{3/4} f(f(x))dx=\int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\\ \int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\\ \int_{1/4}^{1/2}[1]dx=\frac14,$$ because: $$f(f(x))=x^3-\frac32x^2+x+\frac14;\\ f(f(1-x))=(1-x)^3-\frac32(1-x)^2+(1-x)+\frac14= -x^3+\frac32x^2-x+\frac34.$$ Also, the property mentioned by RedFloyd is given in the above source: $$\color{blue}{\int_a^b f(x)dx=\int_a^b f(a+b-x)dx \Rightarrow \\ \int_a^b f(f(x))dx=\int_a^b f(f(a+b-x))dx}.$$