Question

help!!!!!
A área entre as retas y=0, x = -5, X = 3 e a função y(x) = x2 – 36 é:
a) 115,49 u.a.
b) 176.13 u.a.
c) 237,33 u.a.
d) 285,72 u.a
e) 333,33 u.a.

Answer 1

Resposta:

Explicação passo-a-passo:

$A=\int\limits^a_b {[0-(x^{2}-36)]} \, dx \\\\onde\\a=3\\e\\b=-5\\\\A=\int\limits^a_b {[-x^{2}+36]} \, dx\\\\\\A=[\dfrac{-x^{3}}{3}+36x]_-_5^3\\\\\\A=[-\dfrac{3^{3}}{3}+36.3]-[-\frac{(-5)^{3}}{3}+36.(-5)]\\\\\\A=[-\dfrac{27}{3}+108]-[-\frac{(-125)}{3}-180]\\\\\\A=[-9+108]-[\frac{125}{3}-180]\\\\\\A=99-[\frac{125-540}{3}]\\\\\\A=99-[\frac{-415}{3}]\\\\\\A=99+\frac{415}{3}\\\\\\A=\frac{297+415}{3}\\\\\\A=\frac{712}{3}\\\\\\A=237,33$