Notes on Sums, Sqrs/Cubes, and Riemann Sums for f(x)=x^2 over [0,1]

Sums and Area Approximations: Key Ideas from the Transcript

  • Core question: What is the sum of the numbers from 1 to n? i.e. the series 1 + 2 + 3 + \cdots + n = \sum_{i=1}^{n} i

  • Derivation (pairing trick) for the sum of the first n integers:

    • Write the sum in forward and reverse order:
    • Forward: 1 + 2 + 3 + \cdots + n
    • Reverse: n + (n-1) + (n-2) + \cdots + 1
    • Add the two expressions term-by-term. Each pair sums to n+1 and there are n pairs, so:
    • 2\sum_{i=1}^{n} i = n(n+1)
    • Therefore, \sum_{i=1}^{n} i = \frac{n(n+1)}{2}

  • Key formulas to memorize (algebraic sums):

    • Sum of the first n integers:
    • \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
    • Sum of squares (to be used next):
    • \sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}
    • Sum of cubes (remark): the cube sum is the square of the linear sum:
    • \sum_{i=1}^{n} i^{3} = \left(\frac{n(n+1)}{2}\right)^{2}

  • Area under f(x) = x^2 on [0,1] as a guiding value:

    • Exact area (the integral): \int_{0}^{1} x^{2}\,dx = \frac{1}{3}
    • This exact value serves as a guide to check rough estimates from finite sums.

  • Riemann sums as a bridge from algebra to calculus:

    • Consider f(x) = x^2 on [0,1], with a partition into n subintervals of width \Delta x = \frac{1}{n}.
    • Right-endpoint sum (R_n):
    • The right endpoint in each subinterval is x_i = \frac{i}{n} for i = 1,2,…,n.
    • Rn = \sum{i=1}^{n} f(xi) \Delta x = \sum{i=1}^{n} \left(\frac{i}{n}\right)^2 \cdot \frac{1}{n} = \frac{1}{n^{3}} \sum_{i=1}^{n} i^{2}
    • Left-endpoint sum (L_n):
    • Left endpoint in i-th subinterval is x_i^{L} = \frac{i-1}{n}, i = 1,…,n.
    • Ln = \sum{i=1}^{n} f\left(\frac{i-1}{n}\right) \Delta x = \sum{i=1}^{n} \left(\frac{i-1}{n}\right)^{2} \cdot \frac{1}{n} = \frac{1}{n^{3}} \sum{i=0}^{n-1} i^{2}

  • Computing R_n using the sum of squares formula:

    - Using \sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6},

    R_n = \frac{1}{n^{3}} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2n^{3} + 3n^{2} + n}{6\,n^{3}} = \frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^{2}}

    • Therefore, as n \to \infty, R_n \to \frac{1}{3} which matches the true area.

  • Computing L_n similarly:

    - Since \sum_{i=0}^{n-1} i^{2} = \frac{(n-1)n(2n-1)}{6},

    L_n = \frac{1}{n^{3}} \cdot \frac{(n-1)n(2n-1)}{6} = \frac{2n^{3} - 3n^{2} + n}{6\,n^{3}} = \frac{1}{3} - \frac{1}{2n} + \frac{1}{6n^{2}}

    • Thus, L_n \to \frac{1}{3} as n \to \infty as well.

  • Numerical example for n = 4 (to illustrate bounding):

    • Right sum: sum of squares up to i=4 is 30, so
    • R_4 = \frac{1}{4^{3}} \cdot 30 = \frac{30}{64} = \frac{15}{32} = 0.46875
    • Left sum: sum of squares from i=0 to 3 is 14, so
    • L_4 = \frac{1}{4^{3}} \cdot 14 = \frac{14}{64} = \frac{7}{32} = 0.21875
    • True area is \frac{1}{3} \approx 0.3333, so the true value lies between L4 and R4.
    • In general, Rn overestimates and Ln underestimates for convex increasing functions like x^2 on [0,1].

  • The big takeaway: the true area equals the limit of the Riemann sums as the number of rectangles grows:

    • \int{0}^{1} x^{2} \, dx = \lim{n \to \infty} Rn = \lim{n \to \infty} L_n = \frac{1}{3}

  • Conceptual connections and implications:

    • This demonstrates the bridge from algebra (finite sums) to calculus (definite integrals) via limits of sums.
    • The Fundamental Theorem of Calculus (FTC) is invoked as the principle that connects antiderivatives to definite integrals and justifies using integrals (and the table of integrals) once we know the antiderivative.
    • The phrase "FTC license to use the table of integrals" reflects the practical shortcut that, once the FTC framework is in place, one can apply standard integral results rather than re-deriving them from first principles.

  • Practical notes for exam preparation:

    • Always memorize:
    • \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
    • \sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}
    • \sum_{i=1}^{n} i^{3} = \left(\frac{n(n+1)}{2}\right)^{2}
    • Be able to set up right-endpoint and left-endpoint Riemann sums for a simple function like f(x) = x^{2} on a closed interval, compute them in terms of the corresponding sums, and show the bounding behavior as n\to\infty.
    • Use the limit of Rn (and Ln) to verify the definite integral value, reinforcing the concept that the integral is the limit of Riemann sums.

  • Final synthesis (as emphasized in the transcript):

    • The activity of slicing the area into rectangles ("slice and dice") is a concrete, visual method to approximate area and to motivate the definition of the definite integral.
    • The results converge to the exact area as the number of rectangles grows, which is the core idea behind the definition of the integral and its connection to antiderivatives via the FTC.

  • Quick references to the transcript's key phrases:

    • "The sum of the numbers from one to n" corresponds to the formula \sum_{i=1}^{n} i = \frac{n(n+1)}{2}.
    • "The cube should be very simple to memorize"
    • leads to the identity \sum_{i=1}^{n} i^{3} = \left(\frac{n(n+1)}{2}\right)^{2} (sum of cubes equals the square of the sum of integers).
    • "The area under the curve is the limit of the right-hand sum" and the link to FTC.
    • "License to use the table of integrals" refers to FTC as the bridge to applying standard integral results.