Calculus: Integrals I

Sums

Before diving into integrals, we'll need to first make a short study of sums. The sum of two terms is usually written as $a+b$. Big whoop, everyone already knows that. But what about the sum of a lot of numbers? We could write $a+b+c+d+\ldots$, although that'd get a bit tiring. What if we had a compact way to express such a thing? Well guess what. We do! Surprise surprise. Capital letter sigma, $\Sigma$ is used to denote the sum of a set of terms:

$$\displaystyle\sum\limits_{i=1}^{N} i = 1 + 2 + 3 + \ldots + N$$

The variable $i$, known as the index, is used as a counter variable and ranges from $i$ to $N$, where $N$ is some fixed integer. The sum then takes the sum of all the integers from $1$ to $N$.

Sums can be more complicated than the above example. Any summable mathematical expression can be used instead of just plain old $i$. Furthermore, the index need not start at 1. Consider the following algebraic expression:

$$\displaystyle\sum\limits_{i=-2}^{4} x^i = x^{-2} + x^{-1} + 1 + x + x^2 + x^3 + x^4$$

A yet more complicated use of summation notation involves enumerating the values in some sequence of numbers. In this instance, the index variable is used to select which element from the sequence to include in the sum. For example, if $P_i$ denotes the $i$th prime number, then we can express the sum of the first 5 prime numbers as follows:

$$\displaystyle\sum\limits_{i=1}^{5} P_i = 2 + 3+5 +7 + 11$$

Sums are a key component of understanding integrals, not to mention mathematics in general.

Problems

  1. Evaluate the following sums:

    1. $\displaystyle\sum\limits_{i=0}^{10} i$
    2. $\displaystyle\sum\limits_{i=-20}^{20} i$
    3. $\displaystyle\sum\limits_{i=-1}^{5} 2i$
    1. $\displaystyle\sum\limits_{i=0}^{10} i = 1+2+3+4+5+6+7+8+9+10 = 55$
    2. $\displaystyle\sum\limits_{i=-20}^{20} i = -20 + (-19) + (-18) + \ldots + 18 + 19 + 20 = 0$
    3. $\displaystyle\sum\limits_{i=-1}^{5} 2i = -2 + 0 + 2 + 4 +6 +8 + 10 = 28$
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  2. Evaluate the following sums:

    1. $\displaystyle\sum\limits_{i=1}^{5} i^2$
    2. $\displaystyle\sum\limits_{i=1}^{4} i^3$
    3. $\displaystyle\sum\limits_{i=1}^{4} i^i$
    1. $\displaystyle\sum\limits_{i=1}^{5} i^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 \\ \displaystyle\sum\limits_{i=1}^{5} i^2 = 1 + 4 + 9 + 16 + 25 \\ \displaystyle\sum\limits_{i=1}^{5} i^2 = 45 \\ $
    2. $\displaystyle\sum\limits_{i=1}^{4} i^3 = 1^3 + 2^3 + 3^3 + 4^3 \\ \displaystyle\sum\limits_{i=1}^{4} i^3 = 1 + 8 + 27 + 64 \\ \displaystyle\sum\limits_{i=1}^{4} i^3 = 100 \\ $
    3. $\displaystyle\sum\limits_{i=1}^{4} i^i = 1^1 + 2^2 + 3^3 + 4^4 \\ \displaystyle\sum\limits_{i=1}^{4} i^i = 1 + 4 + 27 + 256 \\ \displaystyle\sum\limits_{i=1}^{4} i^i = 288 \\ $
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  3. Let $\left\{a_i\right\}_{i=1}^{\infty}$ denote the sequence of numbers where $a_i$ is the $i$th prime number. That is, $a_1=2$, $a_2=3$, $a_3=5$, and so on. Evaluate the following sums:

    1. $\displaystyle\sum\limits_{k=1}^{5} 2a_k+1$
    2. $\displaystyle\sum\limits_{k=1}^{5} \sin\left(\dfrac{a_k\pi}{2}\right)$
    3. $\displaystyle\sum\limits_{k=1}^{5} \cos\left(\dfrac{a_k\pi}{4}\right)$
    1. $\displaystyle\sum\limits_{k=1}^{5} 2a_k+1 = (2(2)+1) + (2(3)+1) + (2(5)+1) + (2(7)+1) + (2(11)+1) \\ \displaystyle\sum\limits_{k=1}^{5} 2a_k+1 = 5 + 7 + 11 + 15 + 23 \\ \displaystyle\sum\limits_{k=1}^{5} 2a_k+1 = 61 \\ $
    2. $\displaystyle\sum\limits_{k=1}^{5} \sin\left(\dfrac{a_k\pi}{2}\right) = \sin\left(\dfrac{2\pi}{2}\right) + \sin\left(\dfrac{3\pi}{2}\right) + \sin\left(\dfrac{5\pi}{2}\right) + \sin\left(\dfrac{7\pi}{2}\right) + \sin\left(\dfrac{11\pi}{2}\right) \\ \displaystyle\sum\limits_{k=1}^{5} \sin\left(\dfrac{a_k\pi}{2}\right) = 0 + (-1) + 1 + (-1) + (-1)\\ \displaystyle\sum\limits_{k=1}^{5} \sin\left(\dfrac{a_k\pi}{2}\right) = -2 \\ $
    3. $\displaystyle\sum\limits_{k=1}^{5} \cos\left(\dfrac{a_k\pi}{4}\right) = \cos\left(\dfrac{2\pi}{4}\right) + \cos\left(\dfrac{3\pi}{4}\right) + \cos\left(\dfrac{5\pi}{4}\right) + \cos\left(\dfrac{7\pi}{4}\right) + \cos\left(\dfrac{11\pi}{4}\right) \\ \displaystyle\sum\limits_{k=1}^{5} \cos\left(\dfrac{a_k\pi}{4}\right) = 0 + \left(\dfrac{-1}{\sqrt{2}}\right) + \left(\dfrac{-1}{\sqrt{2}}\right) + \left(\dfrac{1}{\sqrt{2}}\right) + \left(\dfrac{-1}{\sqrt{2}}\right) \\ \displaystyle\sum\limits_{k=1}^{5} \cos\left(\dfrac{a_k\pi}{4}\right) = -\sqrt{2} \\ $
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  4. Let $f(x) = \dfrac{4}{x^2+1}$. Evaluate $\displaystyle\sum\limits_{k=-1}^{3} f(x)$

    $\displaystyle\sum\limits_{k=-1}^{3} f(x) = f(-1) + f(0) + f(1) + f(2) + f(3) \\ \displaystyle\sum\limits_{k=-1}^{3} f(x) = \dfrac{4}{(-1)^2+1} + \dfrac{4}{0^2+1} + \dfrac{4}{1^2+1} + \dfrac{4}{2^2+1} + \dfrac{4}{3^2+1} \\ \displaystyle\sum\limits_{k=-1}^{3} f(x) = \dfrac{4}{1+1} + \dfrac{4}{1} + \dfrac{4}{1+1} + \dfrac{4}{4+1} + \dfrac{4}{9+1} \\ \displaystyle\sum\limits_{k=-1}^{3} f(x) = \dfrac{4}{2} + \dfrac{4}{1} + \dfrac{4}{2} + \dfrac{4}{5} + \dfrac{4}{10} \\ \displaystyle\sum\limits_{k=-1}^{3} f(x) = \dfrac{46}{5} \\ $
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  5. Let $f(x)=2x+1$ and let $g(x) = 3x$. Evaluate $\displaystyle\sum\limits_{i=2}^{7} f(i)g(i)$

    $\displaystyle\sum\limits_{i=2}^{7} f(i)g(i) = \displaystyle\sum\limits_{i=2}^{7} (2i+1)(3i) \\ \displaystyle\sum\limits_{i=2}^{7} f(i)g(i) = \displaystyle\sum\limits_{i=2}^{7} 6i^2+3i \\ \begin{align}\displaystyle\sum\limits_{i=2}^{7} f(i)g(i) = & \,\, (6(2)^2+3(2)) + (6(3)^2+3(3)) + (6(4)^2+3(4)) + (6(5)^2+3(5)) \,\, + \\ & \,\, (6(6)^2+3(6)) + (6(7)^2+3(7)) \end{align} \\ \displaystyle\sum\limits_{i=2}^{7} f(i)g(i) = (6(4)+6) + (6(9)+9) + (6(16)+12) + (6(25)+15) + (6(36)+18) + (6(49)+21)\\ \displaystyle\sum\limits_{i=2}^{7} f(i)g(i) = (24+6) + (54+9) + (96+12) + (150+15) + (216+18) + (294+21)\\ \displaystyle\sum\limits_{i=2}^{7} f(i)g(i) = 915\\ $

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  6. Show that $\displaystyle\sum\limits_{i=1}^{n} 1 = n$.

    There really isn't much to show here. The sum is just 1 added to 1 $n$ times, which is known under the more familiar term counting:

    $\displaystyle\sum\limits_{i=1}^{n} 1 = \underbrace{1+1+\ldots+1}_{n} \\ \displaystyle\sum\limits_{i=1}^{n} 1 = n(1) \\ \displaystyle\sum\limits_{i=1}^{n} 1 = n \\ $

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  7. Show that $\displaystyle\sum\limits_{i=1}^{n} i = \dfrac{n(n+1)}{2}$.

    Hint: Look at the first and last terms of the sum.

    Double hint: Try using 2 sums.

    Gauss figured this out when he was a child. That's why he's famous and we're not:

    $\displaystyle\sum\limits_{i=1}^{n} i = 1 + 2 + 3 + \ldots + n \\ \displaystyle\sum\limits_{i=1}^{n} i + \displaystyle\sum\limits_{i=1}^{n} i = 1 + 2 + 3 + \ldots + n + \displaystyle\sum\limits_{i=1}^{n} i\\ \displaystyle\sum\limits_{i=1}^{n} i + \displaystyle\sum\limits_{i=1}^{n} i = \left(1 + 2 + 3 + \ldots + n\right) + \left(1 + 2 + 3 + \ldots + n\right) \\ 2\displaystyle\sum\limits_{i=1}^{n} i = \left(1 + 2 + 3 + \ldots + n\right) + \left(1 + 2 + 3 + \ldots + n\right) \\ 2\displaystyle\sum\limits_{i=1}^{n} i = \left(1 + 2 + 3 + \ldots + n\right) + \left(n + (n-1) + (n-2) + \ldots + 1 \right) \\ 2\displaystyle\sum\limits_{i=1}^{n} i = (1+n) + (2+(n-1)) + (3+(n-2)) + \ldots + (n+1) \\ 2\displaystyle\sum\limits_{i=1}^{n} i = (1+n) + (n+1) + (n+1) + \ldots + (n+1) \\ 2\displaystyle\sum\limits_{i=1}^{n} i = n(n+1) \\ \displaystyle\sum\limits_{i=1}^{n} i = \dfrac{n(n+1)}{2} \\$

    Isn't that cool? It's pretty cool.

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  8. Show that $\displaystyle\sum\limits_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}$.

    Hint: Use induction.

    Proof by induction.

    Base case:

    $\displaystyle\sum\limits_{i=1}^{1} i = \dfrac{1(1+1)(2(1)+1)}{6} \\ \displaystyle\sum\limits_{i=1}^{1} i = \dfrac{1(2)(3)}{6} \\ \displaystyle\sum\limits_{i=1}^{1} i = \dfrac{6}{6} \\ \displaystyle\sum\limits_{i=1}^{1} i = 1 \\ $

    Inductive Step:

    Given $\displaystyle\sum\limits_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}$, we must show that $\displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{(n+1)(n+1+1)(2(n+1)+1))}{6}$.

    $\displaystyle\sum\limits_{i=1}^{n+1} i^2 = (n+1)^2 + \displaystyle\sum\limits_{i=1}^{n} i^2 \\ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = (n+1)^2 + \dfrac{n(n+1)(2n+1)}{6} \\ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{6(n+1)^2}{6} + \dfrac{n(n+1)(2n+1)}{6} \\ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{6(n+1)^2+n(n+1)(2n+1)}{6} \\ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{(n+1) (6(n+1)+n(2n+1))}{6} \\ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{(n+1) (6n+6+2n^2+n))}{6} \\ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{(n+1) (2n^2+7n+6))}{6} \\ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{(n+1) 2(n^2+3.5n+3))}{6} \\ $

    We can factor this quadratic equation by using the quadratic formula:

    $n = \dfrac{-3.5 \pm \sqrt{3.5^2-4(1)(3)}}{2(1)} \\ n = \dfrac{-3.5 \pm \sqrt{12.25-12}}{2} \\ n = \dfrac{-3.5 \pm \sqrt{0.25}}{2} \\ n = \dfrac{-3.5 \pm 0.5}{2} \\ n = \dfrac{-4}{2} = -2 \\ n = \dfrac{-3}{2} = -1.5 \\ $

    Now we substitute the factored quadratic back into the original equation:

    $ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{(n+1) 2(n+2)(n+1.5)))}{6} \\ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{(n+1) (n+2)(2n+3))}{6} \\ \displaystyle\sum\limits_{i=1}^{n+1} i^2 = \dfrac{(n+1)(n+1+1)(2(n+1)+1))}{6} \\ $

    $$\tag*{$\blacksquare$}$$

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