Tutorial Week 4

Partial Fraction Decomposition (PFD)

Partial fraction decomposition is a way to simplify rational integrals into a way that is easier to work with by seperating the rational expression into its partial fractions.

Q1: Integrate \(\int \frac{x^4+2x^3-3x+4}{x(x-3)(x^2+1)} \; dx\).

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Riemann Sums and Sigma Notation

Sigma Notation

We use sigma (\(\Sigma\)) notation as a way to express sums of numbers.

With \(\Sigma_{n=3}^5n^2\), we sum from \(n = 3\) to \(n = 5\) inclusive, giving \(\Sigma_{n=3}^5n^2 = 3^2 + 4^2 + 5^2\).

With sums, we also have a few useful formulas, those being:

  • \(\Sigma_{k = m}^n ca_k = c\Sigma_{k = m}^n a_k\)

  • \(\Sigma_{k = m}^n a_k + b_k = \Sigma_{k = m}^n a_k + \Sigma_{k = m}^n b_k\)

  • \(\Sigma_{k = 1}^n 1 = n\)

  • \(\Sigma_{k = 1}^n n = \frac{n(n+1)}{2}\)

  • \(\Sigma_{k = 1}^n n^2 = \frac{n(n+1)(2n+1)}{6}\)

  • \(\Sigma_{k = 1}^n n^3 = \frac{n^2(n + 1)^2}{4}\)

Riemann Sums

Riemann sums are used to approximate the area under a function by using rectangles.

To define a riemann sum from on the interval [a, b] for f(x), we need:

  • \(\Delta x = \frac{b - a}{n}\)

  • \(x_k = a + k\Delta x\) for \(k = 0, 1, 2, ... , n\)

  • \(x^\ast_k\) is a point in \([x_{k-1}, x_k]\)

The riemann sum is \(\Sigma_{k=1}^n f(x_n^\ast)\Delta x\).

We have two common types of riemann sums, those being the left riemann sum and the right riemann sum.

The left riemann sum is defined with \(x^\ast_k = x_{k-1}\) \(L_n = \Sigma_{k=1}^n f(x_{k-1})\Delta x = \Sigma_{k=1}^n f(a + (k-1)\Delta x)\Delta x\)

The right riemann sum is defined with \(x^\ast_k = x_{k}\) \(R_n = \Sigma_{k=1}^n f(x_{k})\Delta x = \Sigma_{k=1}^n f(a + k\Delta x)\Delta x\)

Q2: Approximate the area under \(10 - x^2\) for \(x \in [0, 1]\) using the right Riemann sum over four intervals.

We first need to find the following:

  • \(\Delta x = \frac{b - a}{n} = \frac{1 - 0}{4} = \frac{1}{4}\)

  • \(x_k = a + k\Delta x = 0 + \frac{k}{4}\) for \(k = 0, 1, 2, ... , n\)

  • \(x^\ast_k = x_{k} = \frac{k}{4}\) since we’re using the right Riemann sum.

  • \(\Sigma_{k=1}^n f(x_{k})\Delta x = \Sigma_{k=1}^n f(\frac{k}{4})\Delta x\)

Then the right riemann sum is

\(R_n = \Sigma_{k=1}^n f(\frac{k}{4})\Delta x = (\frac{1}{4})((10 - (\frac{1}{4})^2) + (10 - (\frac{2}{4})^2) + (10 - (\frac{3}{4})^2) + (10 - 1^2))\)