This companion doc works through every problem from the Unit 6 homework sheet. Each example shows the full setup and arithmetic so you can see where every number comes from. Use this alongside the concepts doc to check your own attempts.
Reference table for Problems 1 through 5:
x | -3 | -1 | 1 | 3 | 5 | 7 | 9 |
|---|---|---|---|---|---|---|---|
h(x) | 5 | 2 | -3 | -7 | -2 | 6 | 11 |
Interval: [-3, 1]. Number of subintervals: n = 2.
Delta x = (1 - (-3)) / 2 = 4 / 2 = 2.
Partition points: x_0 = -3, x_1 = -1, x_2 = 1.
For a left Riemann sum, use the left endpoint of each subinterval:
Subinterval [-3, -1]: height = h(-3) = 5
Subinterval [-1, 1]: height = h(-1) = 2
Left sum = delta x * [h(-3) + h(-1)] = 2 * [5 + 2] = 2 * 7 = 14
Interval: [-3, 9]. Number of subintervals: n = 3.
Delta x = (9 - (-3)) / 3 = 12 / 3 = 4.
Partition points: x_0 = -3, x_1 = 1, x_2 = 5, x_3 = 9.
For a right Riemann sum, use the right endpoint of each subinterval:
Subinterval [-3, 1]: height = h(1) = -3
Subinterval [1, 5]: height = h(5) = -2
Subinterval [5, 9]: height = h(9) = 11
Right sum = 4 * [-3 + (-2) + 11] = 4 * 6 = 24
Interval: [-3, 9]. Number of subintervals: n = 3.
Delta x = 4 (same partition as Problem 2).
Partition points: -3, 1, 5, 9.
Midpoints of each subinterval:
Subinterval [-3, 1]: midpoint = (-3 + 1) / 2 = -1
Subinterval [1, 5]: midpoint = (1 + 5) / 2 = 3
Subinterval [5, 9]: midpoint = (5 + 9) / 2 = 7
All three midpoints (-1, 3, 7) appear in the table. This is required for the method to work with tabulated data.
Midpoint sum = 4 * [h(-1) + h(3) + h(7)] = 4 * [2 + (-7) + 6] = 4 * 1 = 4
Interval: [-3, 3]. Number of subintervals: n = 3.
Delta x = (3 - (-3)) / 3 = 6 / 3 = 2.
Partition points: x_0 = -3, x_1 = -1, x_2 = 1, x_3 = 3.
Trapezoidal sum = (delta x / 2) * [h(x_0) + 2 h(x_1) + 2 h(x_2) + h(x_3)]
= (2 / 2) * [h(-3) + 2 h(-1) + 2 h(1) + h(3)]
= 1 * [5 + 2(2) + 2(-3) + (-7)]
= 1 * [5 + 4 + (-6) + (-7)]
= -4
Interval: [-3, 9]. Number of subintervals: n = 6.
Delta x = (9 - (-3)) / 6 = 12 / 6 = 2.
Partition points: -3, -1, 1, 3, 5, 7, 9 (every value in the table).
Trapezoidal sum = (delta x / 2) * [h(-3) + 2 h(-1) + 2 h(1) + 2 h(3) + 2 h(5) + 2 h(7) + h(9)]
= (2 / 2) * [5 + 2(2) + 2(-3) + 2(-7) + 2(-2) + 2(6) + 11]
= 1 * [5 + 4 + (-6) + (-14) + (-4) + 12 + 11]
= 1 * 8
= 8
Integral of 2x sin(x) from 0 to pi, n = 4.
Delta x = (pi - 0) / 4 = pi/4.
Partition points: 0, pi/4, pi/2, 3pi/4, pi.
For a right Riemann sum, use the right endpoints: pi/4, pi/2, 3pi/4, pi.
Evaluate f(x) = 2x sin(x) at each:
f(pi/4) = 2(pi/4) sin(pi/4) = (pi/2)(sqrt(2)/2) = pi sqrt(2) / 4 ≈ 1.1107
f(pi/2) = 2(pi/2) sin(pi/2) = pi(1) = pi ≈ 3.1416
f(3pi/4) = 2(3pi/4) sin(3pi/4) = (3pi/2)(sqrt(2)/2) = 3 pi sqrt(2) / 4 ≈ 3.3322
f(pi) = 2(pi) sin(pi) = 2 pi (0) = 0
Right sum = (pi/4) * [1.1107 + 3.1416 + 3.3322 + 0]
= (pi/4) * 7.5845
≈ 0.7854 * 7.5845
≈ 5.956
Integral of (e^2)(x^2) from -2 to 10, n = 4.
Note: e^2 is a constant (approximately 7.389). The integrand is (e^2)(x^2).
Delta x = (10 - (-2)) / 4 = 12 / 4 = 3.
Partition points: -2, 1, 4, 7, 10.
Evaluate f(x) = (e^2)(x^2) at each partition point:
f(-2) = e^2 * 4 = 4 e^2 ≈ 29.556
f(1) = e^2 * 1 = e^2 ≈ 7.389
f(4) = e^2 * 16 = 16 e^2 ≈ 118.224
f(7) = e^2 * 49 = 49 e^2 ≈ 361.945
f(10) = e^2 * 100 = 100 e^2 ≈ 738.906
Trapezoidal sum = (delta x / 2) * [f(-2) + 2 f(1) + 2 f(4) + 2 f(7) + f(10)]
= (3/2) * [4 e^2 + 2 e^2 + 32 e^2 + 98 e^2 + 100 e^2]
= (3/2) * 236 e^2
= 354 e^2
≈ 2615.73
Reference table for Problems 8 and 9:
x | -2 | 0 | 1 | 3 | 5 | 8 | 9 |
|---|---|---|---|---|---|---|---|
P(x) | 5 | 8 | 2 | -4 | -1 | 2 | 5 |
Interval: [-2, 9], three subintervals, midpoint Riemann sum.
With seven data points and three subintervals, the spacing is unequal. The standard approach: group the data into three subintervals so the interior data point of each group serves as the evaluation point.
Grouping:
Subinterval [-2, 1], width 3, evaluation point x = 0 (the interior table value)
Subinterval [1, 5], width 4, evaluation point x = 3
Subinterval [5, 9], width 4, evaluation point x = 8
Each subinterval uses its own width (these are unequal, so you cannot factor out a single delta x):
Midpoint sum = 3 * P(0) + 4 * P(3) + 4 * P(8)
= 3(8) + 4(-4) + 4(2)
= 24 + (-16) + 8
= 16
Same table as Problem 8. Interval: [-2, 9], six subintervals.
With seven data points and six subintervals, each consecutive pair of x-values forms one subinterval. The widths are unequal:
[-2, 0]: width 2
[0, 1]: width 1
[1, 3]: width 2
[3, 5]: width 2
[5, 8]: width 3
[8, 9]: width 1
For unequal subintervals, compute each trapezoid separately:
Trap sum = (2/2)[P(-2) + P(0)] + (1/2)[P(0) + P(1)] + (2/2)[P(1) + P(3)] + (2/2)[P(3) + P(5)] + (3/2)[P(5) + P(8)] + (1/2)[P(8) + P(9)]
= 1[5 + 8] + 0.5[8 + 2] + 1[2 + (-4)] + 1[(-4) + (-1)] + 1.5[(-1) + 2] + 0.5[2 + 5]
= 13 + 5 + (-2) + (-5) + 1.5 + 3.5
= 16
Q: A function f is continuous and increasing on [2, 10]. Is a left Riemann sum an overestimate or an underestimate of the integral from 2 to 10 of f(x) dx?
A: Underestimate. When f is increasing, the left endpoint of each subinterval gives a height that is lower than every other point in that subinterval, so each rectangle sits below the curve.
Q: You are given four equally spaced data points at x = 0, 3, 6, 9 with f(0) = 1, f(3) = 4, f(6) = 2, f(9) = 5. Compute the trapezoidal sum for the integral from 0 to 9.
A: Delta x = 3. Trap sum = (3/2)[f(0) + 2 f(3) + 2 f(6) + f(9)] = (3/2)[1 + 8 + 4 + 5] = (3/2)(18) = 27.
Q: Why does the trapezoidal sum equal the average of the left and right Riemann sums?
A: The left sum uses f(x_0) through f(x_{n-1}) and the right sum uses f(x_1) through f(x_n). Averaging them gives (1/2)[f(x_0) + 2 f(x_1) + ... + 2 f(x_{n-1}) + f(x_n)] times delta x, which is exactly the trapezoidal formula (after factoring the 1/2 into delta x / 2).
Q: A concave-down function g is defined on [1, 5]. Is the trapezoidal approximation an overestimate or underestimate?
A: Underestimate. When the curve is concave down, the straight-line segments connecting the data points lie below the curve, so the trapezoids miss area that sits above them.
Q: You have six subintervals and want to compute a midpoint Riemann sum from a table. How many data points do you need, and which ones do you use?
A: You need 13 data points (or at minimum the 6 midpoint values plus the 7 boundary values). For the midpoint sum itself, you use only the 6 midpoint values. Each is the function value at the centre of its subinterval.
Q: The integral from 0 to 8 of f(x) dx is approximated with four subintervals. What is delta x, and what are the partition points?
A: Delta x = (8 - 0) / 4 = 2. Partition points: 0, 2, 4, 6, 8.