Use five rectangles to approximate the area under the curve.

Answer 1

Note: the answer to 1) was already provided by Harold. And 2) can be done very similarly.

3) This is done with sigma notation, from i = 1 to i = n, limit of n--->infinity. The letter n represents the number of intervals. The width of each rectangle will be delta x = 20/n. If you use the right endpoint (the left could also be used, with slight adjustments to fit) for each interval, x_i = 20i/n, and the value of the given function will be

y_i = 20(20i/n) - (20i/n)².

The area = lim n-->infinity sigma from i = 1 to i = n [y_i x 20/n]

= ditto [20(20i/n) - (20i/n)²](20/n)

= 8000 [(1/n²)sigma i - (1/n³)sigma i²]

* = 8000 [(1/n²)(n)(n + 1)/2 - (1/n³)(n)(n + 1)(2n + 1)/6]

= 8000 [(1/2 + 1/2n) - (1/3 + 1/2n + 1/6n²)]

= 8000 (1/2 - 1/3)

= 8000/6

= 4000/3, approximately 1333.33

These summations for i and i² are known, and proofs are in textbooks. The first can be proved similarly to the famous method used by Gauss to add all the integers from 1 to 100.

Answer 2

This Numerical Methods approximation problem is best tackled using Riemann Sums.

We will use Left Sums (LHS) since they are the easiest.

1)

Solve: ∫ 20x-x² dx from a=0 to b=20

This is approximated by P₀(x) = lim ∑ f(x_i) Δx for i = 1 to 5 with n=5 intervals

The base of the rectangle is Δx.

Δx = (b-a)/n = (20-0)/5 = 4

We will evaluate 5 rectangles at the following x values:

x = a + Δx*i where i = 0 to n-1 or 0 to 4

x₀ = a + Δx*0 = 0 + 4*0 = 0

x₁ = a + Δx*1 = 0 + 4*1 = 4

x₂ = a + Δx*2 = 0 + 4*2 = 8

x₃ = a + Δx*3 = 0 + 4*3 = 12

x₄ = a + Δx*4 = 0 + 4*4 = 16

The area of each rectangle is base times height, or:

A = b*h = Δx * f(x_i) using the x_i values above.

Rectangle #1 @ x₀: A=b*h = Δx * f(x₀) = Δx * f(0) = 4 * (20*0-0²) = 4 * (0) = 0

Rectangle #2 @ x₁: A=b*h = Δx * f(x₁) = Δx * f(4) = 4 * (20*4-4²) = 4 * (80-16) = 4*64 = 256

Rectangle #3 @ x₂: A=b*h = Δx * f(x₂) = Δx * f(8) = 4 * (20*8-8²) = 4 * (160-64) = 4*96 = 384

Rectangle #4 @ x₃: A=b*h = Δx * f(x₃) = Δx * f(12) = 4 * (20*12-12²) = 4 * (240-144) = 4*96 = 384

Rectangle #5 @ x₄: A=b*h = Δx * f(x₄) = Δx * f(16) = 4 * (20*16-16²) = 4 * (320-256) = 4*64 = 256

NOTE: Notice the symmetry.

The answer, which is an approximation, is the sum of all 5 rectangles.

A = 0 + 256 + 384 + 384 + 256 = 1,280

2)

Similar to 1) above, but with n=10

3)

Let's use a proper integral since as n → ∞, Δx → dx

Solve: ∫ 20x-x² dx from a=0 to b=20

F(x) = (20x²)/2 - (x³)/3

F(x) = 10x² - (1/3)x³

Evaluating from 0 to 20 gives:

A = F(20) - F(0) = [10(20)² - (1/3)(20)³] - [10(0)² - (1/3)(0)³]

= [4000 - 8000/3] - [0] = 4000/3 ≅ 1,333,3

This answer is exact.

Notice how the answer increases in accuracy as the interval, n, increases.

Do a Google search for "Harold’s Calculus Notes Cheat Sheet" and see page 7 for the formulas.