Consider the function \(f(x)=6x^2-10x+13\). Is there another function, \(F(x)\), in which \(f\) is the derivative of? In other words, \(\frac{d}{dx}\left(F(x)\right)=f(x)\)? When such a function exists, we call \(F\) an antiderivative of \(f\).
The good news is that we can take \(f\) one step at a time. The first term is \(6x^2\), a power function. Trying to think backwards, the \(x^2\) has to come from \(x^3\). By the Power Rule, \(\frac{d}{dx}\left(x^3\right)=3x^2\), but the coefficient is a \(6\), not a \(3\). However, \(6=2\cdot3\), so it must be that \(2x^3\) is the term that has \(6x^2\) as the derivative:
$$\frac{d}{dx}\left(2x^3\right)=6x^2\implies F(x)=2x^3$$
In practice, to compensate for that multiplier of \(3\), we can divide it out and get the same result:
$$f(x)=6x^2\rightarrow\left\{\begin{array}{l} \implies6\left(x^2\right) \\ \implies 6\left(\frac{1}{2+1}x^{2+1}\right) \\ \implies6\left(\frac{1}{3}x^3\right) \\ \implies\frac{6}{3}x^3 \\ \implies2x^3 \end{array}\right\}\rightarrow F(x)=2x^3$$
Using this process on each term of \(f\) leads to \(F(x)=2x^3-5x^2+13x\). There is just one problem with \(F\), does it have one more term?
Since \(\frac{d}{dx}\left(c\right)=0\) for any constant \(c\), there might be a contant at the end: \(F(x)=2x^3-5x^2+13x+10\), or \(F(x)=2x^3-5x^2+13x-27\). This constant could be anything, and gives many possibilites for an antiderivative. Using a captial \(C\) describes all the antiderivatives:$$F(x)=2x^3-5x^2+13x+C$$
Before getting into specific rules for things, it gets a bit cumbersome to use the \(F\) as the notation for antiderivatives. Recall the notation: \(\frac{d}{dx}\left(f(x)\right)\) means to take the derivative of \(f\) with respect to the variable \(x\).
The notation for antiderivatives if called the indefinite integral and is written: \(\int f(x)dx\)
Note there are not limits of integration like in the definite integral, \(\int_{a}^{b}f(x)dx\). That is how to distinguish between a problem finding area and a problem finding the antiderivative. The example above using this notation would look like: $$\int\left(6x^2-10x+13\right)dx=2x^3-5x+13x+C$$
The above example leads to a more general rule to "work backwards" from the derivative Power Rule:
$$\int ax^n dx=\frac{a}{n+1}x^{n+1}$$ where \(n\ne-1\) and \(C\) is an arbitraty constant
Since \(\frac{d}{dx}\left(\sin(ax)\right)=a\cos(ax)\) it follows that \(\int\left(\cos(ax)\right)dx=\large{\frac{1}{a}}\normalsize\sin(ax)+C\). The antiderivatives of the other trigonometric functions follow similarly:
BE WARY OF NEGATIVES!!!
$$\int\left(\cos(ax)\right)dx=\frac{1}{a}\sin(ax)+C$$
$$\int\left(\sin(ax)\right)dx=-\frac{1}{a}\cos(ax)+C$$
$$\int\left(\sec^2(ax)\right)dx=\frac{1}{a}\tan(ax)+C$$
$$\int\left(\csc^2(ax)\right)dx=-\frac{1}{a}\cot(ax)+C$$
$$\int\left(\sec(ax)\tan(ax)\right)dx=\frac{1}{a}\sec(ax)+C$$
$$\int\left(\csc(ax)\cot(ax)\right)dx=-\frac{1}{a}\csc(ax)+C$$
Recall that \(\frac{d}{dx}\left(b^x\right)=b^x\cdot \ln(b)\). It is important to understand that \(\ln(b)\) is a number, and just like above, to compensate for this number we must divide it out.
$$\int b^x dx=\frac{1}{\ln(b)}b^x+C$$ where \(b>0,\;b\ne1\) and \(C\) is an arbitraty constant
SPECIAL CASE:
since \(\frac{d}{dx}\left(e^x\right)=e^x\), it follows \(\int e^xdx=e^x+C\).
Recall in the Power Rule above, the condition of \(n\ne1\). If \(n=1\), the function looks like \(f(x)=x^-1\). Attempting to apply the Power Rule here would give
$$\int x^{-1}dx=\frac{1}{-1+1}x^{-1+1}+C\implies\frac{1}{\color{red}{0}}x^0+C$$
giving an undefined value. Now \(f(x)=x^{-1}\) can be rewritten into \(f(x)=\large{\frac{1}{x}}\). What function then has \(f\) as the derivative?
The function \(F(x)=\ln x\) does the trick since \(\frac{d}{dx}\left(\ln x\right)=\large{\frac{1}{x}}\). Therefore, \(\int\frac{1}{x}dx=\ln|x|+C\).
Here are the problems in the video. Attempt these first on your own, and then watch the video.
Any equation involving an unknown function and its derivative is called a differential equation. For example, \(f^\prime(x)=4x-7\), in which we describe ALL antiderivatives using \(f(x)=2x^2-7x+C\). A natural question is, how can the value of \(C\) be determined?
To achieve this, one more piece of information is needed called an intitial value. For this example, consider the initial condition that \(f(10)=114\). Now plug in the initial condition to find \(C\):
$$\begin{array}{l} f(\color{blue}{10})=\color{green}{114} \\ \implies \color{green}{114}=2(\color{blue}{10})^2-7(\color{blue}{10})+C \\ \implies 114=200-70+C \\ \implies 114=130+C \\ \implies -16=C \end{array}$$
Now instead of describing ALL the antiderivatives with \(C\), we have found the one solution to the differential equation \(f^\prime(x)=4x-7\) with initial value condition \(f(10)=114\). The solution is \(f(x)=2x^2-7x-16\color{green}{\checkmark}\).