-
Write down a formula for an antiderivative
of each of these functions.
Note you can always check your antiderivative
by taking its derivative.
\( a'(x) = 124x^{123}\)\( b'(x) = x^{421}\)\( c'(x) = x^4+x^3+x^2+x+1 \)\( d'(x) = \left(1-x^6\right)^2\)\( f'(x) = 2\sqrt[3]{x} + \frac{3}{x^6} \)\( g'(x) = \frac{1}{\sqrt[3]{x}} \)\( h'(x) = \cos(x)\)\( j'(x) = \sin(x) + 1\)\( k'(x) = 4\sec(x)\tan(x)\)\( m'(x) = \csc^2(x)\)\( n'(x) = 3\cos\!\left(3x\right)\)\( p'(x) = 2x\cos\!\left(x^2\right)\)\( q'(x) = x^7\cos\!\left(x^8\right)\)
-
A cyclist is wearing a heart rate monitor
while competing in a fifty-mile bike race.
They check the monitor, which tracks
their heart rate measured in beats-per-minute (bpm),
occasionally throughout the race.
First, briskly sketch a scatter plot of this data with your independent variable \(t\) measured in minutes since the start of the race.
Minutes since Start 0 10 33 58 70 101 120 Heart Rate 66 73 80 88 97 68 87 - Estimating, how many times did their heart beat in those first \(10\) minutes of the race? What about the first \(33\) minutes of the race? What about the first \(60\) minutes of the race?
- Assume that, for the period between any two times that the cyclist checks their heart rate, their heart rate is between the rates the monitor reports. (E.g. between minutes \(58\) and \(70,\) their heart rate stays between \(88\) bpm and \(97\) bpm.) Under this assumption, can you come up with an upper-bound for the number of times their heart beat during the race? I.e. can you find a number \(M\) and say for certain their heart beat no more than \(M\) times during the race? What about a lower-bound?
- Now let’s assume something stronger: assume that, for the period between any two times that the cyclist checks their heart rate, their heart rate can be accurately described by a linear function of time. (E.g. between minutes \(58\) and \(70,\) their heart rate is increasing at a constant rate of \(\frac{3}{4}\) bpm-per-minute.) Now since we are assuming we know their heart rate at any time throughout the race, we should be able to determine a exact number for the number of times their heart beat. How do we determine this number?
- We often denote the \(n^\text{th}\) derivative of a function \(f\) as \(f^{(n)}.\) For example, \(f^{(13)}(x)\) is the thirteenth derivative of \(f\) — this is better notation than \(f'''''''''''''(x).\) Given that \({f^{(5)}(x) = 12x+1}\) what’s a possible formula for \(f(x)?\) What’s a possible formula for \(f(x)\) if we also require that \(f^{(3)}(2)=21?\)
-
Without relying on technology, determine the value of each of these sums.
\( \displaystyle \sum_{n=1}^{4} \frac{1}{n} \)\( \displaystyle \sum_{n=0}^{5} \frac{2}{3}n+2 \)\( \displaystyle \sum_{n=1}^{9} (-1)^n \)\( \displaystyle \sum_{n=-9}^{10} n^3 \)\( \displaystyle \sum_{n=79}^{300} n \)\( \displaystyle \sum_{n=58}^{75} n^2 \)
-
Express each of the following sums in sigma (\(\Sigma\)) summation notation.
\(2 + 4 + 6 + 8 + 10 + 12\)\(1 + 3 + 5 + 7\)\(\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{16}{81}\)\(6 - 7 + 8 - 9 + 10 - 11 + 12\)\(1 + 4 + 9 + 16 + \dotsb + 529\)\(\frac{\sqrt{3}}{2}+\frac{1}{2}+0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}-\frac{1}{2}-0+\frac{1}{2}+\frac{\sqrt{3}}{2}+1\)
Challenges
- Consider the following sum where the upper-limit is some variable \(M.\) \[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} +\frac{1}{32} +\frac{1}{64} +\frac{1}{128} +\frac{1}{256} +\frac{1}{512} +\frac{1}{1024} +\frac{1}{2048} +\dotsb +\frac{1}{2^M} \] What value does this sum approach as \(M\) approaches \(\infty?\)
- Consider the following sum where the upper-limit is some variable \(M.\) \[\frac{4}{1} -\frac{4}{3} +\frac{4}{5} -\frac{4}{7} +\frac{4}{9} -\frac{4}{11} +\frac{4}{13} -\frac{4}{15} +\frac{4}{17} -\frac{4}{19} +\frac{4}{21} -\frac{4}{23} +\dotsb+(-1)^{M}\frac{4}{2M+1} \] What value does this sum approach as \(M\) approaches \(\infty?\)
-
Recall the classic proof that \( 1+2+3+4+\dotsb+n = \frac{1}{2}n(n+1)\,. \)
-
Using a similar “number arrangement” argument, prove that \[ 1^2+2^2+3^2+4^2+\dotsb+n^2 = \frac{n(n+1)(2n+1)}{6}\,. \]
- Can you extend your proof to show that \( 1^3+2^3+3^3+4^3+\dotsb+n^3 = \frac{n^2(n+1)^2}{4}\,? \)
-