Question

Determine a área total entre a curva y = x^2-3x-10 e o eixo x no intervalo [ -2, 8]

Answer 1

Resposta:

$Area=|A_{1}|+A_{2}\\\\\\\displaystyle A_{1}=\int\limits^5_{-2} {(x^2-3x-10)} \, dx=\left.(\frac{x^3}{3}-\frac{3x^{2}}{2}-10x)\right|\limits_{-2}^{5}=\\\\\\=(\frac{5^3}{3}-\frac{3.5^{2}}{2}-10.5)-[\frac{(-2)^3}{3}-\frac{3.(-2)^{2}}{2}-10.(-2)]=\\\\\\=(\frac{125}{3}-\frac{75}{2}-50)-[-\frac{8}{3}-6+20]=-\frac{343}{6}$

$\displaystyle A_{2}=\int\limits^8_{5} {(x^2-3x-10)} \, dx=\left.(\frac{x^3}{3}-\frac{3x^{2}}{2}-10x)\right|\limits_{5}^{8}=\\\\\\=(\frac{8^3}{3}-\frac{3.8^{2}}{2}-10.8)-[\frac{5^3}{3}-\frac{3.5^{2}}{2}-10.5]=\\\\\\=(\frac{512}{3}-96-80)-[\frac{125}{3}-\frac{75}{2} -50]=\frac{243}{6} \\\\\\Area=|-\frac{343}{6} |+\frac{243}{6}=\frac{293}{3}$

Answer 2

Resposta:

Explicação passo a passo: