I will do the exact value part, I don't know how your text or your notes
define "Left Sum, Right Sum, Midvalue Sum, and Trapezoid Rule methods"
∫ x^3 dx from 3 to 21
= [(1/4) x^4] from 3 to 21
= (1/4)(21^4) - (1/4)(3^4)
= 194481/4 - 81/4
= 48600 units^2
Consider the integral:
21
∫ x^3 dx
3
-Use Left Sum, Right Sum, Midvalue Sum, and Trapezoid Rule methods to approximate this integral with n = 3
-Find the exact value of this integral
2 answers
As for the approximations, you have n=3, so you want three intervals, each of width (21-3)/3 = 6. That gives you
x0=3, x1=9, x2=15, x=321
Left Sum:
2
∑ 6f(xk)
k=0
Right Sum:
3
∑ 6f(xk)
k=1
Midpoint Sum:
2
∑ 6f((xk+xk+1)/2)
k=0
Trapezoid Sum:
2
∑ 6(f(xk)+f(xk+1
))/2
k=0
x0=3, x1=9, x2=15, x=321
Left Sum:
2
∑ 6f(xk)
k=0
Right Sum:
3
∑ 6f(xk)
k=1
Midpoint Sum:
2
∑ 6f((xk+xk+1)/2)
k=0
Trapezoid Sum:
2
∑ 6(f(xk)+f(xk+1
))/2
k=0
1 answer