# $\int_{-1/2}^{3/2} xf(3x^2 - 2x^3) \,dx = 2 \int_{0}^{1} xf(3x^2 -2x^3) \,dx$ proof

I have faced a confusing maths problem that goes :

$$f : [-1/2, 3/2] \to\mathbb R$$ is a continuous function

Prove that :

$${\int_{-1/2}^{3/2} x f(3x^2 -2x^3) \,dx =2 \int_0^1 xf(3x^2 - 2x^3) dx}$$

I thought about using the formula : $${\int_a^b\ f(x) dx = \frac{a+b}2\int_a^b f(x)dx }$$

and the formula : $${\int_a^b\ f(x) dx =\int_a^b f(a+ b -x)dx }$$

Nevertheless, they both seem not to help. Even trying the hints mentioned in the comments is fruitless.

• Hint: Try plotting the function $3x^2-2x^3$ and then look for a suiting set of substitutions. Oct 17, 2023 at 16:48
• Just a guess if you have $a(x) = [g(x)+g(-x)]/2$ and $b(x)=[g(x)-g(-x)]/2$, Then $g(x)= a(x)+b(x)$ and $\int_{-c}^c b(x) dx =0, \ \int_{-c}^c a(x) dx = 2\int_0^{c} a(x)dx$ combined with $x=p(u)$ might do it. The doubling of second integral and the partial symmetry in the first version is suggestive. Oct 17, 2023 at 17:49
• I second @MaximilianJanisch comment. Oct 17, 2023 at 20:24
• Title's support MathJax. Nov 11, 2023 at 12:38
• @Salem: Ah OK. I think you need some more calculus level for this problem (However i don't discard to be wrong about it). Regards. Dec 10, 2023 at 14:32

HINT.- An idea leading to a proof. Denote $$(*)$$ the proposed equality.
$$(1)$$The equality $$(*)$$ is true for all monomial $$f(x)=a_nx^n$$.
In fact $$\displaystyle a_n\int_{-1/2}^{3/2} x(3x^2-2x^3)^ndx=2a_n\int_0^1x^{1+2n}(3-2x)^ndx$$ we can verify this for all integer $$n$$. (For example, discarding WLOG the constant coefficient $$a_n$$, we have in both integrals $$(*)$$ the values $$1,0.7,0.405623$$ for exponents $$0,1,5$$ respectively). This probably can be proven by induction on $$n$$.
$$(2)$$ Because of linearity of integrals, it follows that $$(*)$$ is true for all polynomial $$f(x)$$.
$$(3)$$Now, for Weirstrass approximation theorem, every continuous real function $$f(x)$$ defined on the compact $$[-1/2,3/2]$$ is uniformly limit of polynomials $$P_n(x)$$ and each of them satisfy $$(*)$$.