Question

Dados os graficos encontre as areas: ME AJUDEM!!!!

Answer 1

18)

a) y = -3x⁷ + 4


Integrando a função no intervalo (0,1):

A = $\int\limits^1_0 {(-3x^7 \ + \ 4)} \, dx$

A = $(\frac{-3x^{7+1}}{7+1} \ + \ 4x)]^1_0$

A = $(\frac{-3x^{8}}{8} \ + \ 4x)]^1_0$


Substituindo "x":

A = $(\frac{-3.(1)^{8}}{8} \ + \ 4.(1)) - (\frac{-3.(0)^{8}}{8} \ + \ 4.(0))$

A = $(\frac{-3.1}{8} \ + \ 4) - (\frac{-3.0}{8} \ + \ 0)$

A = $(\frac{-3}{8} \ + \ 4) - (\frac{0}{8} \ + \ 0)$

A = $(\frac{-3}{8} \ + \ 4) - (0 \ + \ 0)$

A = $\frac{-3}{8} \ + \ 4$


MMC:

A = $\frac{-3 \ + \ 32}{8}$

A = $\frac{29}{8}$

A = 3,625



b)

f(x) = -x  ---> y = -x
g(x) = x² - 1 ---> y = x² - 1

Igualando as funções:

y = y
-x = x² - 1
x² + x - 1 = 0

Determinando o intervalo da função:

Δ = (1)² - 4.(1).(-1)
Δ = 1 + 4
Δ = 5

x' = (-1 + √5)/2
x' = (-1 + 2,236068)/2
x' = 1,236/2
x' = 0,618034


x" = (-1 - √5)/2
x" = (-1 - 2,236068)/2
x" = -3,236/2
x" = -1,618034

 
Logo:

$A = \int\limits^{0,618034}_{-1,618034} {(-x) - (x^2 - 1)} \, dx$


Integrando:

$A = (\frac{-x^{1+1}}{1+1}) - (\frac{x^{2+1}}{2+1} - x)]^{0,618034}_{-1,618034}$

$A = (\frac{-x^{2}}{2} - (\frac{x^{3}}{3} - x)]^{0,618034}_{-1,618034}$

$A = (\frac{-x^{2}}{2} - \frac{x^{3}}{3} + x)]^{0,618034}_{-1,618034}$


Substituindo "x":

$A = (\frac{-(0,618034)^{2}}{2} - \frac{(0,618034)^{3}}{3} + 0,618034) - (\frac{-(-1,618034)^{2}}{2} - \frac{(-1,618034)^{3}}{3} + (-1,618034))$

$A = (\frac{-0,381966}{2} - \frac{0,236068}{3} + 0,618034) - (\frac{-2,618034}{2} - \frac{-4,236068}{3} - 1,618034)$

$A = (\frac{-0,381966}{2} - \frac{0,236068}{3} + 0,618034) - (\frac{-2,618034}{2} + \frac{4,236068}{3} - 1,618034)$

$A = (-0,190983 - 0,078689 + 0,618034) - (-1,309017 + 1,412022 - 1,618034)$

A = 0,348362 - (-1,515029)
A = 0,348362 + 1,515029
A = 1,863391