The formula for the area of a trapezoid is 1/2(b_1 + b_2)h where h = delta (x)
Therefore, A = 1/2(4+k)2 + 1/2(k+8)2 + 1/2(8+12)2 = 54
Solving that equation yields k = 10.
The function f is continuous on the closed interval [0,6] and has values that are given in the table below.
x |0|2|4|6
f(x)|4|K|8|12
The trapezoidal approximation for(the integral):
6
S f(x) dx
1
found with 3 subintervals of equal length is 52. What is the value of K?
How do you do this with the formula
b-a/(2n) * f(x0) + 2(fx1) + f(xn)?
1 answer
1 answer