187 lines
11 KiB
Markdown
187 lines
11 KiB
Markdown
# Antiderivatives
|
|
An antiderivative is useful when you know the rate of change, and you want to find a point from that rate of change
|
|
|
|
> A function $F$ is said to be an *antiderivative* of $f$ if $F'(x) = f(x)$
|
|
## Notation
|
|
The collection of all antiderivatives of a function $f$ is referred to as the *indefinite integral of $f$ with respect to $x$*, and is denoted by:
|
|
$$ \int f(x) dx $$
|
|
## Examples
|
|
> Find the antiderivative of the function $y = x^2$
|
|
|
|
1. We know that to find the derivative of the above function, you'd multiply by the exponent ($2$), and subtract 1 from the exponent.
|
|
2. To perform this operation in reverse:
|
|
1. Add 1 to the exponent
|
|
2. Multiply by $\dfrac{1}{n + 1}$
|
|
3. This gives us an antiderivative of $\dfrac{1}{3}x^3$
|
|
4. To check our work, work backwards.
|
|
5. The derivative of $\dfrac{1}{3}x^3$ is $\dfrac{1}{3} (3x^2)$
|
|
6. $= \dfrac{3}{3} x^2$
|
|
|
|
|
|
## Formulas
|
|
|
|
| Differentiation Formula | Integration Formula |
|
|
| ----------------------------------------------------- | -------------------------------------------------------- |
|
|
| $\dfrac{d}{dx} x^n = nx^{x-1}$ | $\int x^n dx = \dfrac{1}{n+1}x^{n+1}+ C$ for $n \ne -1$ |
|
|
| $\dfrac{d}{dx} kx = k$ | $\int k \space dx = kx + C$ |
|
|
| $\dfrac{d}{dx} \ln \|x\| = \dfrac{1}{x}$ | <br>$\int \dfrac{1}{x}dx = \ln \|x\| + C$ |
|
|
| $\dfrac{d}{dx} e^x = e^x$ | <br>$\int e^x dx = e^x + C$ |
|
|
| $\dfrac{d}{dx} a^x = (\ln{a}) a^x$ | $\int a^xdx = \ln \|x\| + C$ |
|
|
| $\dfrac{d}{dx} \sin x = \cos x$ | $\int \cos(x) dx = \sin (x) + C$ |
|
|
| $\dfrac{d}{dx} \cos x = -\sin x$ | $\int \sin(x)dx = \sin x + C$ |
|
|
| $\dfrac{d}{dx} \tan{x} = \sec^2 x$ | $\int \sec^2(x)dx = \tan(x) + C$ |
|
|
| $\dfrac{d}{dx} \sec x = \sec x \tan x$ | $\int sec^2(x) dx = \sec(x) + C$ |
|
|
| $\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}}$ | $\int \sec(x) \tan(x) dx = \sec x + C$ |
|
|
| $\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2}$ | $\int \dfrac{1}{\sqrt{1+x^2}}dx = \tan^{-1}x + C$ |
|
|
| $\dfrac{d}{dx} k f(x) = k f'(x)$ | $\int k*f(x)dx = k\int f(x)dx$ |
|
|
| $\dfrac{d}{dx} f(x) \pm g(x) = f'(x) \pm g'(x)$ | $\int (f(x) \pm g(x))dx = \int f(x) dx \pm \int g(x) dx$ |
|
|
# Area Under a Curve
|
|
The area under the curve $y = f(x)$ can be approximated by the equation $\sum_{i = 1}^n f(\hat{x_i})\Delta x$ where $\hat{x_i}$ is any point on the interval $[x_{i - 1}, x_i]$, and the curve is divided into $n$ equal parts of width $\Delta x$
|
|
|
|
Any sum of this form is referred to as a Reimann Sum.
|
|
|
|
To summarize:
|
|
- The area under a curve is equal to the sum of the area of $n$ rectangular subdivisions where each rectangle has a width of $\Delta x$ and a height of $f(x)$.
|
|
# Definite Integrals
|
|
Let $f$ be a continuous function on the interval $[a, b]$. Divide $[a, b]$ into $n$ equal parts of width $\Delta x = \dfrac{b - a}{n}$ . Let $x_0, x_1, x_2, \cdots, x_3$ be the endpoints of the subdivision.
|
|
|
|
The definite integral of $f(x)$ with respect to $x$ from $x = a$ to $x = b$ can be denoted:
|
|
$$ \int_{a}^b f(x) dx $$
|
|
|
|
And __can__ be defined as:
|
|
$$ \int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i = 1}^n f(x_i)\Delta x$$
|
|
|
|
$f(x_i)$ is the *height* of each sub-interval, and $\Delta x$ is the change in the *x* interval, so $f(x_i) \Delta x$ is solving for the area of each sub-interval.
|
|
|
|
- If your function is always positive, then the value of a definite integral is the area under the curve.
|
|
- If the function is always negative, then the value of a definite integral is the area above the curve to zero.
|
|
- If the function has both positive and negative values, the output is equal to the area above the curve minus the area below the curve.
|
|
|
|
## Examples
|
|
> Find the exact value of the integral $\int_0^1 5x \space dx$
|
|
|
|
Relevant formulas:
|
|
$$ \sum_{i = 1}^n = \dfrac{(n)(n + 1)}{2} $$
|
|
$$ \Delta x = \dfrac{1 - 0}{n} = \dfrac{1}{n}$$$$ x_i = 0 + \Delta xi + \dfrac{1}{n} \cdot i$$
|
|
1. $\int_0^1 5x \space dx = \lim_{n \to \infty} \sum_{i=1}^n 5(x_i) \cdot \Delta x$
|
|
2. $= \lim_{n \to \infty} \sum_{i=1}^n 5(\frac{1}{n} \cdot i) \cdot \frac{1}{n}$
|
|
3. $= \lim_{n \to \infty} \sum_{i = 1}^n \dfrac{5}{n^2}\cdot i$
|
|
4. $= \lim_{n \to \infty} \dfrac{5}{n^2} \sum_{i = 1}^n i$
|
|
5. $= \lim_{x \to \infty} \dfrac{5}{n^2} \cdot \dfrac{n(n + 1)}{2}$
|
|
6. $= \lim_{n \to \infty} \dfrac{5n^2 + 5n}{2n^2}$
|
|
7. $= \dfrac{5}{2}$
|
|
|
|
# Properties of Integrals
|
|
1. $\int_a^a f(x)dx = 0$ - An integral with a domain of zero will always evaluate to zero.
|
|
2. $\int_b^a f(x)dx = -\int_a^b f(x) dx$ - The integral from $a \to b$ is equal to the integral from $-(b\to a)$
|
|
3. $\int_a^b cf(x) dx = c \int_a^b f(x) dx$ - A constant from inside of an integral can be moved outside of an integral
|
|
4. $\int_a^b f(x) \pm g(x) dx = \int_a^b f(x) dx \pm \int_a^b g(x)dx$ - Integrals can be distributed
|
|
5. $\int_a^c f(x)dx = \int_a^b f(x)dx + \int_b^c f(x)dx$ - An integral can be split into two smaller integrals covering the same domain, added together.
|
|
|
|
# Averages
|
|
To find the average value of $f(x)$ on the interval $[a, b]$ is given by the formula:
|
|
|
|
Average = $\dfrac{1}{b-a} \int_a^b f(x)dx$
|
|
|
|
# The Fundamental Theorem of Calculus
|
|
1. Let $f$ be a continuous function on the closed interval $[a, b]$ and let $F$ be any antiderivative of $f$, then:
|
|
$$\int_a^b f(x) dx = F(b) - F(a)$$
|
|
2. Let $f$ be a continuous function on $[a, b]$ and let $x$ be a point in $[a, b]$.
|
|
$$ F(x) = \int_a^x f(t)dt \Rightarrow F'(x) = f(x) $$
|
|
This basically says that cancelling out the derivative from $a$ to $x$ can be done by taking the derivative of that equation. with respect to $x$.
|
|
$$ \dfrac{d}{dx} \int_a^{g(x)} f(t) dt = f(g(x)) * g'(x)* $$
|
|
## Examples
|
|
> Finding the derivative of an integral
|
|
$$ \dfrac{d}{dx} \int_2^{7x} \cos(t^2) dt = cos((7x)^2) * 7 = 7\cos(49x^2)$$
|
|
> Finding the derivative of an integral
|
|
$$ \dfrac{d}{dx}\int_0^{\ln{x}}\tan(t) = \tan(\ln(x))*\dfrac{1}{x} $$
|
|
> $x$ and $t$ notation *(note: the bar notation is referred to as "evaluated at")*
|
|
$$ F(x) = \int_4^x 2t \space dt = t^2 \Big|_4^x = x^2 - 16$$
|
|
> $x$ in top and bottom
|
|
$$ \dfrac{d}{dx} \int_{2x}^{3x} \sin(t) dt = \dfrac{d}{dx} -\cos(t)\Big|_{2x}^{3x} = \dfrac{d}{dx} (-\cos(3x) + cos(2x) = 3\sin(3x) - 2\sin(2x) $$
|
|
|
|
# The Mean Value Theorem for Integrals
|
|
If $f(x)$ is continuous over an interval $[a, b]$ then there is at least one point $c$ in the interval $[a, b]$ such that:
|
|
$$f(c) = \dfrac{1}{b-a}\int_a^bf(x)dx $$
|
|
This formula can also be stated as $\int_a^b f(x)dx = f(c)(b-a)$
|
|
This theorem tells us that a continuous function on the closed interval will obtain its average for at least one point in the interval.
|
|
|
|
# U-Substitution
|
|
## Formulas
|
|
- $\int k {du} = ku + C$
|
|
- $\int u^n du = \frac{1}{n+1}u^{n+1} + C$
|
|
- $\int \frac{1}{u} du = \ln(|u|) + C$
|
|
- $\int e^u du = e^u + C$
|
|
- $\int \sin(u) du = -\cos(u) + C$
|
|
- $\int \cos(u) du = \sin(u) + C$
|
|
- $\int \dfrac{1}{\sqrt{a^2 - u^2}} du = \arcsin(\frac{u}{a}) +C$
|
|
- $\int \dfrac{1}{a^2+u^2}du = \dfrac{1}{a} \arctan(\frac{u}{a}) + C$
|
|
- $\int \dfrac{1}{u\sqrt{u^2 - a^2}} du = \dfrac{1}{a}arcsec(\dfrac{|u|}{a}) + C$
|
|
|
|
# Length of a Curve
|
|
## Review of the Mean Value Theorem
|
|
If $f$ is a continuous function on the interval $[a, b]$ and differentiable on $(a, b)$, then there exists a number $c$ in the interval $(a, b)$ such that:
|
|
|
|
$$ f'(c) = \dfrac{f(b) - f(a)}{b - a} $$
|
|
|
|
This also implies that for some $c$ in the interval $(a, b)$:
|
|
$$ f(b) - f(a) = f'(c)(b-a) $$
|
|
|
|
## Intuitive Approach
|
|
Given that we divide a curve into $n$ sub-intervals, and we can find the location of the right endpoint of each interval.
|
|
|
|
With a series of points on a curve we can find the distance between each point.
|
|
|
|
As we increase $n$, the precision of which the curve is estimated increases.
|
|
|
|
This means that:
|
|
$$ \text{length of a curve} = \lim_{n \to \infty} \sum_{i=1}^{n}(\text{length of line segment)}$$
|
|
Using the distance formula, we know that the length of the line segment can be found with:
|
|
$$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
|
|
1. So the entire equation is:
|
|
$$ \text{length of a curve} = \lim_{n \to \infty} \sum_{i=1}^{n}(\sqrt{(x_i - x_{i-1})^2 + (y_i - y_{i-1})^2}) $$
|
|
This can also be described as:
|
|
$$ \text{length of a curve} = \lim_{n \to \infty} \sum_{i=1}^{n}(\sqrt{(\Delta x)^2 +(\Delta y)^2}) $$
|
|
2. Using the mean value theorem:
|
|
$$ \lim_{n \to \infty} \sum_{i = 1}^n)\sqrt{\Delta x^2 + (f(x_i) - f(x_{i-1}))i^2} $$
|
|
$$ \lim_{n \to \infty} \sum_{i=1}^n \sqrt{\Delta x ^2 + (f'(x_{\hat{i}}))(x_i - x_{i-1})^2}$$
|
|
3. Factoring out $\Delta x$
|
|
$$ \lim_{n \to \infty} \sum_{i=1}^n \sqrt{\Delta x^2(1 + f'(x_{\hat{i}}))}$$
|
|
4. Moving $\Delta x$ out of the root
|
|
|
|
$$ \lim_{n \to \infty} \sum_{i=1}^n \sqrt{(1 + f'(x_{\hat{i}}))} \Delta x$$
|
|
5. As an integral:
|
|
$$ L =\int_a^b \sqrt{1 + f'(x)^2} dx$$
|
|
## Examples
|
|
> Find the length of the curve $y = -\frac{5}{12}x + \frac{7}{12}$ from the point $(-1, 1)$
|
|
|
|
1. $L = \int_{-1}^8 \sqrt{1 + (-\frac{5}{12})^2} dx$
|
|
2. $= \int_{-1}^8 \sqrt{1 + \frac{25}{144}} dx$
|
|
3. = $\int_{-1}^8 \sqrt{\frac{169}{144}}dx$
|
|
4. $= \int_{-1}^8 \frac{13}{12} dx$
|
|
5. $\frac{13}{12} x \Big| _{-1}^8$
|
|
|
|
> Find the distance from the point ${\frac{1}{2}, \frac{49}{48}}$ to the point $(5, \frac{314}{15})$ along the curve $y = \dfrac{x^4 - 3}{6x}$.
|
|
> *note*: The complete evaluation of this problem is more work than typically required, and is only done for demonstration purposes.
|
|
1. $y' = \dfrac{4x^3(6x) - (x^4 + 3)6}{36x^2}$: Find the derivative of the curve using the quotient rule
|
|
2. $= \dfrac{18x^4 - 18}{36x^2}$: Simplify
|
|
3. $= \dfrac{18(x^4 - 1)}{18(2x^2)}$: Factor out $18$
|
|
4. $= \dfrac{x^4 - 1}{2x^2}$: Factor out $18$ again
|
|
5. $L = \int_{1/2}^5 \sqrt{1 + (\dfrac{4x-1}{2x^2})^2}dx$ : Use the length formula
|
|
6. $= \int_{1/2}^5 \sqrt{1 + \dfrac{x^8 - 2x^4 + 1}{x^4}} dx$: Apply the $^2$
|
|
7. $= \int_{1/2}^5 \sqrt{\dfrac{4x^4 + x^8 -2x^4 + 1}{4x^4}}dx$: Set $1 = \dfrac{4x^4}{4x^4}$ and add
|
|
8. $= \int_{1/2}^5 \sqrt{\dfrac{x^8 + 2x^4 + 1}{4x^4}}dx$: Factor the numerator
|
|
9. $= \int_{1/2}^5 \sqrt{\dfrac{(4x+1)^2}{4x^4}}dx$ : Get rid of the square root
|
|
10. = $\int_{1/2}^5 \dfrac{x^4 + 1}{2x^2}dx$: Move the constant $\frac{1}{2}$ outside of the integral
|
|
11. $= \frac{1}{2}\int_{1/2}^5 \dfrac{x^4 + 1}{x^2}$: Rewrite to remove the fraction
|
|
12. $= \frac{1}{2} \int_{1/2}^5 (x^4 + 1)(x^{-2})dx$: distribute
|
|
13. $= \frac 1 2 \int_{1/2}^5 (x^2 - x^{-2})dx$: Find the indefinite integral
|
|
14. $= \dfrac{1}{2} (\frac{1}{3}x^3 - x^-1)\Big|_{1/2}^5$ : Plug and chug
|
|
15. $= (\frac{125}{6} - \frac{1}{10}) - (\frac{1}{48} - 1)$
|
|
16. $=(\frac{5000}{240} - \frac{24}{240}) - (\frac{5}{240} - \frac{240}{240})$
|
|
|
|
> Find the length of the curve $y = \sqrt{1 - x^2}$
|
|
1. $y$ has a domain of $[-1, 1]$
|
|
2. $y' = \dfrac{1}{2}(1-x^2)^{-1/2}(-2x)$
|
|
3. $= -\dfrac{x}{\sqrt{1 - x^2}}$
|
|
4. $L = \int_{-1}^1 \sqrt{1 + (-\frac{x}{\sqrt{1-x^2}})}$ |