Question

Suponha que $f(x)=ax^2+bx+c$ e $\int\limits^{1}_{-1} {x^2f(x)} \, dx =2$
Encontre a função $f(x)$ que resule num valor mínimo de $\int\limits^{1}_{-1} {[f(x)]^2} \, dx$

Answer 1

Resposta: f(x) = 5x²

 

Vamos lá. Supondo que f(x) = ax² + bx + c :

 

$\begin{array}{l}\int\limits^{\sf1}_{\sf-1}\sf x^2\,f(x)\,dx=2\\\\\int\limits^{\sf1}_{\sf-1}\sf x^2\,(ax^2+bx+c)\,dx=2\\\\\int\limits^{\sf1}_{\sf-1}\sf(ax^4+bx^3+cx^2)\,dx=2\end{array}$

 

Pelo teorema fundamental do cálculo, no qual

• $\int\limits^{\sf b}_{\sf a}\sf f(x)\,dx=F(x)\big|^{\sf b}_{\sf a}=F(b)-F(a)$

, segue que:

 

$\begin{array}{l}\int\sf(ax^4+bx^3+cx^2)\,dx\,\big|^{1}_{-1}=2\\\\\int\sf ax^4\,dx+\int\sf bx^3\,dx+\int\sf cx^2\,dx\,\big|^{1}_{-1}=2\\\\\sf\dfrac{ax^5}{5}+\dfrac{bx^4}{4}+\dfrac{cx^3}{3}\,\bigg|^{1}_{-1}=2\\\\\sf\dfrac{a(1)^5}{5}+\dfrac{b(1)^4}{4}+\dfrac{c(1)^3}{3}-\bigg(\dfrac{a(-1)^5}{5}+\dfrac{b(-1)^4}{4}+\dfrac{c(-1)^3}{3}\bigg)=2\\\\\sf\dfrac{a}{5}+\dfrac{b}{4}+\dfrac{c}{3}-\bigg(-\dfrac{a}{5}+\dfrac{b}{4}-\dfrac{c}{3}\bigg)=2\end{array}$

$\begin{array}{l}\sf\dfrac{a}{5}+\dfrac{b}{4}+\dfrac{c}{3}+\dfrac{a}{5}-\dfrac{b}{4}+\dfrac{c}{3}=2\\\\\sf\dfrac{2a}{5}+\dfrac{2c}{3}=2~(i)\end{array}$

 

Por hora ficamos por isso mesmo. Vamos agora calcular o valor mínimo de f(x) por:

 

$\begin{array}{l}\int\limits^{\sf1}_{\sf-1}\sf[f(x)]^2\,dx=\int\limits^{\sf1}_{\sf-1}\sf(ax^2+bx+c)^2\,dx\\\\=\int\limits^{\sf1}_{\sf-1}\sf(a^2x^4+b^2x^2+c^2+2abx^3+2acx^2+2bcx)\,dx\\\\=\int\limits^{\sf1}_{\sf-1}\sf(a^2x^4+(2ac+b^2)x^2+c^2+2abx^3+2bcx)\,dx\end{array}$

 

Pelo teorema fundamental do cálculo:

 

$\begin{array}{l}\int\limits^{\sf1}_{\sf-1}\sf[f(x)]^2\,dx=\int\sf(a^2x^4+(2ac+b^2)x^2+c^2+2abx^3+2bcx)\,dx\,\big|^{1}_{-1}\\\\=\int\sf a^2x^4\,dx+\int\sf (2ac+b^2)x^2\,dx+\int\sf c^2\,dx+\int\sf2abx^3\,dx+\int\sf2bcx\,dx\,\big|^{1}_{-1}\\\\\sf=\dfrac{a^2x^5}{5}+\dfrac{(2ac+b^2)x^3}{3}+c^2x+\dfrac{2abx^4}{4}+bcx^2\,\bigg|^{1}_{-1}\end{array}$

$\begin{array}{l}\sf=\dfrac{a^2}{5}+\dfrac{2ac+b^2}{3}+c^2+\dfrac{2ab}{4}+bc+\dfrac{a^2}{5}+\dfrac{2ac+b^2}{3}+c^2-\dfrac{2ab}{4}-bc\\\\\sf=\dfrac{2a^2}{5}+\dfrac{4ac+2b^2}{3}+2c^2~(ii)\end{array}$

 

Vejamos.. pela equação (i) anteriormente encontrada, temos:

 

$\begin{array}{l}\sf\dfrac{2a}{5}+\dfrac{2c}{3}=2\\\\\sf\dfrac{6a+10c}{15}=2\\\\\sf6a+10c=15\cdot2\\\\\sf6a+10c=30\end{array}$

 

Sendo assim, vamos tentar inserir esse valor na equação (ii):

 

$\begin{array}{l}\int\limits^{\sf1}_{\sf-1}\sf[f(x)]^2\,dx=\dfrac{2a^2}{5}+\dfrac{4ac}{3}+\dfrac{2b^2}{3}+2c^2\\\\\sf=\dfrac{6a^2+20ac+20b^2+30c^2}{15}\\\\\sf=\dfrac{6a^2+10ac+10ac+20b^2+30c^2}{15}\\\sf=\dfrac{a\overbrace{\sf(6a+10c)}^{\sf30}+10ac+20b^2+30c^2}{15}\\\\\sf=\dfrac{30a+10ac+20b^2+30c^2}{15}\end{array}$

 

Note que quando b = c = 0, temos que:

 

$\begin{array}{l}\sf 6a+10(0)=30\implies 6a+0=30\iff a=\dfrac{30}{6}\iff a=5\end{array}$

 

Portanto,

 

$\begin{array}{l}\int\limits^{\sf1}_{\sf-1}\sf[f(x)]^2\,dx=\dfrac{30(5)+10(5)(0)+20(0)^2+30(0)^2}{15}\\\\\int\limits^{\sf1}_{\sf-1}\sf[f(x)]^2\,dx=\dfrac{30(5)+0+0+0}{15}\\\\\int\limits^{\sf1}_{\sf-1}\sf[f(x)]^2\,dx=2(5)\\\\\int\limits^{\sf1}_{\sf-1}\sf[f(x)]^2\,dx=10\end{array}$

(esse é o valor mínimo.)

 

Ou seja, quando a = 5, b = 0 e c = 0, temos $\int\limits^{\sf1}_{\sf-1}\sf[f(x)]^2\,dx$ como um valor mínimo. Logo, a função que queríamos encontrar é:

 

$\begin{array}{l}\sf f(x)=ax^2+bx+c\\\\\sf f(x)=5x^2+0x+0\\\\\red{\boldsymbol{\sf f(x)=5x^2}}\end{array}$