urgent calculus help (trapezoidal rule)

Question

 

Using 4 equal-width intervals, show that the trapezoidal rule is the average of the upper and lower sum estimates for 

Answer 1

The lower estimate will be the left-hand sum (since x² increasing on [0,4]). The upper will be the right-hand sum. Left and right-hand sums have to do with at which x-values we evaluate the function in order to get the heights of the rectangles so we can sum their areas. The interval is [0,4] so 4 rectangles (and trapezoids) will mean 4 subintervals of = width at [0,1] , [1,2] , [2,3], and [3,4].

The left-hand sum uses the y-values associated with the left endpoints of those subintervals, namely 0, 1, 2, and 3. f(0) = 0 , f(1) = 1, f(2) = 4, and f(3) = 9. So the lower sum is 1·(0 + 1 + 4 + 9) = 14.

The righthand sum uses the righthand endpoints, 1, 2, 3, and 4. f(4) = 16, so upper sum is 1·(1 + 4 + 9 + 16) = 30.

Lastly, the trapezoidal sum can be obtained with this formula S_trap = (b-a)/(2n)·(f(a) + 2f(x₁) + ... + 2f(x_n-1) + f(b)). So S_trap = 1/2(0 + 2 + 8 + 18 + 16) = 22.

It is a fairly trivial proof to show that the trapezoidal sum is always the average of the left- and right-hand sums.

Answer 2

Δx = (b - a)/4 = (4 - 0)/4 = 1; f(x) = x², T₄ = 1/2(0² + 2(1² + 2² + 3²) + 4²) = 22

L₄ = 1(0² + 1² + 2² + 3²) = 14; R₄ = 1(1² + 2² + 3² + 4²) = 30

(14 + 30)/2 = 44/2 = 22. So we prove that T₄ = (L₄ + R₄₎/2

Because f'(x) = 2x > 0 on interval (0,4) then f(x) is increasing and R₄ is upper sum and L₄ is lover sum