Don’t worry, you won’t need to actually use much calculus
Limits are crucial for understanding both mechanics of hypothesis testing, and for understanding derivatives
Derivatives are important for understanding instantaneous rates of changes and for optimization
Integrals are crucial for hypothesis testing
A limit helps us understand the behavior of a function as it approaches a certain point.
It tells us the value that a function gets close to as the input approaches a specific value, even if it doesn’t actually reach that value.
Mathematically, we write:
\[ \lim_{{x \to c}} f(x) = L \]
This means that as x gets closer to c, the value of f(x) approaches L.
For a function \(f(x)\), the statement:
\[ \lim_{{x \to c}} f(x) = L \]
means that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that:
\[ 0 < |x - c| < \delta \implies |f(x) - L| < \epsilon \]
If \(\lim \limits_{x \to c} f(x) = L\)
\[ 0 < |x - c| < \delta \implies |f(x) - L| < \epsilon \]
\(\epsilon\) represents how close we want \(f(x)\) to be to L.
\(\delta\) represents how close x must be to c to guarantee that f(x) is within \(\epsilon\) of L
The limit exists if we can find such a \(\delta\) for any \(\epsilon\)
TL;DR: As x gets very close to x, f(x) gets arbitrarily close to L.
If ( f(x) ) is continuous at ( x = c ), you can find the limit by substituting ( c ) directly into the function: \[ \lim_{{x \to c}} f(x) = f(c) \]
If direct substitution leads to an indeterminate form like \(\frac{0}{0}\), try factoring the numerator and denominator, and then simplify: \[ \lim_{{x \to c}} \frac{{x^2 - 1}}{{x - 1}} = \lim_{{x \to c}} \frac{{(x - 1)(x + 1)}}{{x - 1}} = \lim_{{x \to c}} (x + 1) \]
If direct substitution leads to an indeterminate form, differentiate the numerator and denominator: \[ \lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)} \]
Divide by the highest power of x to solve limits as \(x \to \infty\): \[ \lim_{{x \to \infty}} \frac{x + 1}{x} = \lim_{{x \to \infty}} \left( 1 + \frac{1}{x} \right) = 1 \]
Consider the function:
\[ f(x) = \frac{{x^2 - 1}}{{x - 1}} \]
If we plug in x = 1, we get \(\frac{0}{0}\), but the limit exists!
\[ \frac{{x^2 - 1}}{{x - 1}} = \frac{{(x - 1)(x + 1)}}{{x - 1}} = x + 1 \]
Thus, \(\lim\limits_{x\to 1} f(x) = 2\)
\[ f(x) = \frac{x+1}{x} \]
A derivative represents the rate at which a function is changing at any given point. It’s essentially the slope of the function at that point.
The derivative of \(f(x)\) with respect to \(x\) is defined as:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \]
This formula calculates the instantaneous rate of change of \(f(x)\) as \(\Delta x\) becomes infinitesimally small.
Notice how as \(\Delta x\) gets smaller, we approach the true slope (represented by the thick red line).
For a polynomial function of the form:
\[ f(x) = x^n \]
The derivative with respect to x is:
\[ f'(x) = n x^{n - 1} \]
For a composite function f(x) = g(h(x)) , the derivative with respect to x is:
\[f'(x) = g'(h(x)) \cdot h'(x) \]
For two differentiable functions u(x) and v(x):
\[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) v(x) + u(x) v'(x) \]
An integral represents the accumulation of quantities, such as areas under curves. It’s essentially the inverse operation of differentiation.
The definite integral of \(f(x)\) from \(a\) to \(b\) is written as:
\[ \int_{a}^{b} f(x) \, dx \]
This computes the total accumulation of \(f(x)\) between \(x = a\) and \(x = b\).
To find the area under a curve \(f(x)\) from a to b , we can:
Approximating the Area Under a Curve Using 10 Rectangles
Approximating the Area Under a Curve Using 100 Rectangles
Approximating the Area Under a Curve Using 1000 Rectangles
An antiderivative of a function f(x) is another function F(x) such that:
\[ F'(x) = f(x) \]
In other words, the derivative of F(x) is f(x) .
Antiderivatives are also known as indefinite integrals.
The antiderivative of ( f(x) ) is denoted by:
\[ \int f(x) \, dx = F(x) + C \]
\(\int\) is the integral symbol.
C is the constant of integration, representing all possible antiderivatives.
Find the antiderivative of f(x) = 2x.
Solution:
We are looking for a function F(x) such that F’(x) = 2x.
The antiderivative is:
\[ \int 2x \, dx = x^2 + C \]
Antidifferentiation is the reverse process of differentiation.
Antiderivatives allow us to recover the original function from its derivative.
They are essential in computing definite integrals using the Fundamental Theorem of Calculus.
If f is continuous on [a, b] and F is an antiderivative of f, then:
\[\int_{a}^{b} f(x) dx = F(b) - F(a) \]
Compute the area under \(f(x) = x^{2}\) from x = 0 to x = 3.
Find the antiderivative F(x) :
\[F(x) = \int f(x) dx = \int x^2 dx = \frac{x^3}{3} + C\]
Apply the Fundamental Theorem:
\[\int_{0}^{3} x^2 dx = F(3) - F(0) = \left( \frac{3^3}{3} \right) - \left( \frac{0^3}{3} \right) = \left( \frac{27}{3} \right) - 0 = 9 \]
Imagine you want to model the total tax revenue of a city over the next 5 years.
We would expect revenue to increase as a function of economic growth, inflation and population growth.
Let’s assume that the town has current tax revenues of $10,000,000, and that revenue grows at 7% per year.
The exponential growth formula is
\[ R(t)=R(0) ⋅e^{kt} \]where R(0) is revenue at time t = 0, k is the growth rate and t is the time that has passed.
\[ \text{Total Revenue} = \int_{0}^{5} R(t) dt = \int_{0}^{5} 10,000,000\times e^{.07t}dt \]
We pull the initial revenue out front since it’s a constant which leaves us
\[ \text{Total Revenue} = 10,000,000 \int_{0}^{5}e^{.007t}dt \]
The integral of \(e^{kt}\) with respect to t is \[\frac{1}{K} e^{kt}\]
\[ \text{Total Revenue} = 10,000,000 \times(\frac{e^{.35} - e^{0}}{0.07}) \]
Which ends up coming out to a Total Revenue of
\[ 59,860,000 \]
There are many cases where we want to know to accumulation of some quantity (economic growth, expenses, etc) over time.
In this course, understanding integration and the area under the curve will be crucial to understanding hypothesis testing