Tutorial Week 12

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 \(x^\ast_k = x_{k}\) \(R_n = \Sigma_{k=1}^n f(x_{k})\Delta x = \Sigma_{k=1}^n f(a + k\Delta x)\Delta x\)

Q1: 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

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

Integrals

Integrals are essentially riemann sums with an infinite number of partitions. That is, \(\int_a^b f(x) dx = \lim_{n\to \infty} \Sigma_{k=1}^n f(x^\ast_k) \Delta x\) for the riemann sum over \(x \in [a, b]\). The \(dx\) tells you which variable to integrate, which is useful in multivariable calculus.

Q2: Evaluate \(\int_0^4 2x^2 + x \,dx\) using Riemann sums.

For simplicity, let’s use the right riemann sum.

We have:

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

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

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

So then

\[\begin{split}\begin{aligned} \int_1^5 2x^2 + x dx &= \lim_{n \to \infty} \Sigma_{k=1}^n f(x_{k})\Delta x \\ &= \lim_{n \to \infty} \Sigma_{k=1}^n f(\frac{4k}{n})\Delta x \\ &=\lim_{n \to \infty} \Sigma_{k=1}^n (2(\frac{4k}{n})^2 + \frac{4k}{n})(\frac{4}{n}) \end{aligned}\end{split}\]

Solving for this gives us

\[\begin{split}\begin{aligned} &\lim_{n \to \infty} \Sigma_{k=1}^n (2(\frac{4k}{n})^2 + \frac{4k}{n})(\frac{4}{n}) \\ &= \lim_{n \to \infty} \frac{4}{n}\Sigma^n_{k = 1}2(\frac{4k}{n})^2 + \frac{4k}{n} \\ &= \lim_{n \to \infty} \frac{4}{n}\Sigma^n_{k = 1}\frac{32k^2}{n^2} + \frac{4k}{n} \\ &= \lim_{n \to \infty} \frac{4}{n} ( \frac{32}{n^2} \Sigma^n_{k = 1} k^2 + \frac{4}{n} \Sigma^n_{k = 1} k) \\ &= \lim_{n \to \infty} \frac{4}{n} (\frac{32}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{4}{n} \cdot \frac{n(n+1)}{2} ) \\ &= \lim_{n \to \infty} \frac{128(2n^3+3n^2+n)}{6n^3} + \frac{16(n^2 + n)}{2n^2} \\ &= \text{... skipping a few steps for finding the limit} \\ &= \frac{128 \cdot 2}{6} + \frac{16}{2} \\ &= \frac{128}{3} + 8 \end{aligned}\end{split}\]

Integral Properties and Formulas

  • \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) if \(n \neq -1\)

  • \(\int \frac{1}{x} dx = ln|x| + C\)

  • \(\int e^x dx = e^x + C\)

(and a few more for trigonometric functions)

  • \(\int_a^b f(x) + g(x) dx= \int_a^b f(x) dx+ \int_a^b g(x) dx\)

  • \(\int_a^b kf(x) dx = k \int_a^b f(x) dx\)

  • \(\int_a^a f(x) dx = 0\)

  • \(\int_b^a f(x) dx = -\int_a^b f(x) dx\)

  • \(\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx\)

Fundamental Theorem of Calculus 2 (FTC 2)

With FTC 2, you’re able to evaluate integrals at the endpoints of your interval.

If \(\int f(x) dx = F(x)\), then \(\int_a^b f(x) dx = F(x) |_a^b = F(b) - F(a)\)

Q3: Find \(\int_{0}^{1} 5x + x^2\sqrt[3]{x} \, dx\).

../../_images/5.jpg

Q4: Find \(\int^3_3 \frac{\cos(e^{x})}{x^{2}+e^{x}} \, dx\).

../../_images/11.jpg

Q5: If \(\int_{1}^{5} f(x) + 4 \, dx = 8\), \(\int_1^2 f(x) + x \, dx = 2\), find \(\int_2^5 f(x) \, dx\).

../../_images/31.jpg