Respuesta :
Answer:
The area of the region between the graph of the given function and the x-axis = 25,351 units²
Step-by-step explanation:
Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15
If 'f' is a continuous on [a ,b] then the function
[tex]F(x) = \int\limits^a_b {f(x)} \, dx[/tex]
By using integration formula
[tex]\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c[/tex]
Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]
[tex]\int\limits^6_^-6} (x^{5} + 8 x^{4} + 2 x^{2} + 5 x + 15) )dx[/tex]
On integration , we get
= [tex](\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}[/tex]
[tex]F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)[/tex]
= [tex](\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))[/tex]
After simplification and cancellation we get
= [tex]\frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6[/tex]
on calculation , we get
= [tex]\frac{124,416}{5} + \frac{864}{3} + 180[/tex]
On L.C.M 15
= [tex]\frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}[/tex]
= 25 351.2 units²
Conclusion:-
The area of the region between the graph of the given function and the x-axis = 25,351 units²
