Question

Seja uma função definida por f(x)= mx^2+nx tal que, (imagem abaixo)

Answer 1

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$\sf f(x)=mx^2+nx\\\displaystyle\sf\int_0^1 f(x)~dx=\dfrac{13}{6}\\\displaystyle\sf\int_0^1(mx^2+nx)~dx=\dfrac{1}{3}mx^3+\dfrac{1}{2}nx^2\Bigg|_{0}^1=\dfrac{1}{3}m+\dfrac{1}{2}n\\\sf \dfrac{1}{3}m+\dfrac{1}{2}n=\dfrac{13}{6}\cdot6\implies 2m+3n=13$

$\displaystyle\sf\int_1^2(mx^2+nx)~dx=\dfrac{1}{3}mx^3+\dfrac{1}{2}nx^2\Bigg|_{1}^2\\\sf=\dfrac{1}{3}m\cdot2^3+\dfrac{1}{2}n\cdot2^2-\left[\dfrac{1}{3}m\cdot1^3+\dfrac{1}{2}n\cdot1^2\right]\\\sf =\dfrac{8}{3}m+\dfrac{4}{2}n-\dfrac{1}{3}m-\dfrac{1}{2}n=\dfrac{16m+12n-2m-3n}{6}\\\sf =\dfrac{14m+9n}{6}\\\displaystyle\sf\int_1^2 f(x)~dx=\dfrac{55}{6}\\\sf \dfrac{14m+9n}{\diagup\!\!\!6}=\dfrac{55}{\diagup\!\!\!6}\\\sf 14m+9n=55$

$\begin{cases}\sf2m+3n=13\cdot(-3)\\\sf14m+9n=55\end{cases}\\+\underline{\begin{cases}\sf-6m-\diagdown\!\!\!\!\!9n=-39\\\sf14m+\diagdown\!\!\!\!\!9n=55\end{cases}}\\\sf 8m=16\\\sf m=\dfrac{16}{8}=2\\\sf 2\cdot2+3n=13\\\sf 4+3n=13\\\sf 3n=13-4\\\sf 3n=9\\\sf n=\dfrac{9}{3}=3\\\huge\boxed{\boxed{\boxed{\boxed{\sf m+n=2+3=5\checkmark}}}}$