Question

O valor para integral

Answer 1

Resposta: Letra C) 3288/15

Explicação passo a passo:

Calcular a integral definida:

$\displaystyle\int^{5}_{2}\dfrac{(x^4+x^2)}{3}\,dx=I$

Para tal, aplique o teorema fundamental do cálculo:

$\int^b_a f(x)dx=F(x)\big|^b_a=F(b)-F(a)$

$\therefore$

$\begin{array}{l}I=A\displaystyle\int^{5}_{2}\dfrac{(x^4+x^2)}{3}\,dx=\\\\~\;\!=\displaystyle\int\dfrac{(x^4+x^2)}{3}\,dx\,\bigg|^{5}_{2}=\\\\~\;=\dfrac{1}{3}\displaystyle\int(x^4+x^2)\,dx\,\bigg|^{5}_{2}=\\\\~\;=\dfrac{1}{3}\bigg(\displaystyle\int x^4dx+\int x^2dx\bigg)\bigg|^{5}_{2}=\\\\~\;=\dfrac{1}{3}\bigg(\dfrac{x^{4+1}}{4+1}+\dfrac{x^{2+1}}{2+1}\bigg)\bigg|^{5}_{2}=\\\\~\;=\dfrac{1}{3}\bigg(\dfrac{x^5}{5}+\dfrac{x^3}{3}\bigg)\bigg|^{5}_{2}=\end{array}$

   $\begin{array}{l}~\;\!=\dfrac{1}{3}\bigg(\dfrac{3x^5}{15}+\dfrac{5x^3}{15}\bigg)\bigg|^{5}_{2}=\\\\~\;\!=\bigg(\dfrac{3x^5+5x^3}{45}\bigg)\bigg|^{5}_{2}=\\\\~\;\!=\bigg(\dfrac{3.5^5+5.5^3}{45}\bigg)-\bigg(\dfrac{3.2^5+5.2^3}{45}\bigg)=\\\\~\;\!=\dfrac{3.3125+5.125}{45}-\dfrac{3.32+5.8}{45}=\\\\~\;\!=\dfrac{9375+625}{45}-\dfrac{96+40}{45}=\\\\~\;\!=\dfrac{10000}{45}-\dfrac{136}{45}=\\\\~\;\!=\dfrac{9864}{45}=\\\\~\;\!=\dfrac{3288}{15}\end{array}$

Letra C