Integration

Antiderivation §

If the derivative is the rate of change of a function, then the antiderivative is the accumulated change over time. It’s like an onion.

1. $f^{\prime }(x)$ describes the rate of change of $f(x)$
2. $f(x)$ describes the rate of change of $F(x)$
3. $F(x)$ describes the area under $f(x)$ Both $F(x)$ and $f^{\prime }(x)$ say something about $f(x)$, they’re just in different directions. So, turning a normal function into its antiderivative is going to be like reversing a derivation. Let’s look at a derivation together.

$$\begin{array}{rl}f(x)& =x^3+3x^2+1\\ f^{\prime }(x)& =3x^2+6x\end{array}$$

What is the relationship between $f(x)$ and $f^{\prime }(x)$? Like, if we wanted to manipulate $f^{\prime }(x)$ to turn it back into $f(x)$, what do we do?

$$\begin{array}{rl}3x^2+6x& =x^3+3x^2+1\\ \frac{3(x^2\times x)}{3}+\frac{6\times x}{2}& =x^3+3x^2+1\end{array}$$

But that is incorrect— we’re missing the $+1$ at the end. When we find an antiderivative, we don’t know what that $+1$ is supposed to be, but we know that a constant could have been lost when doing our work. For that reason, we have a handy placeholder for any constant— $C$. We can just add $C$ to make our antiderivative have a “placeholder” for the constant. We will also rewrite our value so it looks better.

$$\begin{array}{rl}{F}^{\prime }{f}^{\prime }(x)& =\frac{3(x^2\times x)}{3}+\frac{(6\times x)}{2}+C\\ & =x^3+3x^2+C\end{array}$$

Note that because $C$ represents any constant, you can theoretically solve for it if you have enough information— for example, if you know what $f(x)$ equals, you can solve for $+C$.

Integration Examples §

$$\begin{array}{r}\int \cos (2x)+6x^2-1dx\\ \frac{1}{2}\sin (2x)+2x^3-x+C\end{array}$$

U-Substitution §

For this we will consider:

$$\begin{array}{r}\int {\cos }^3(x)\times \sin (x)dx\end{array}$$

We want $u$ to be one of our terms with $x$ in it, and we know that $\cos (x)$ becomes $-\sin (x)dx$, which will cancel out with $\sin (x)$ later, so we will choose it.

$$\begin{array}{rlrl}u& =\cos (x)& du& =-\sin (x)dx\\ dx& =\frac{du}{-\sin x}& & \end{array}$$

We can then rewrite our integral and integrate with respect to $du$.

$$\begin{array}{rl}& \int u^3\times \sin (x)\left(\frac{du}{-\sin (x)}\right)=\int -u^{3}du=-\int u^{3}du\\ & -\int u^{3}du=-\frac{u^4}{4}+C\end{array}$$

Then, we substitute back in $u$ to get our answer.

$$-\frac{{\cos }^4(x)}{4}+C$$

Partial Fractional Decomposition §

$$\begin{array}{rl}f(x)& =\frac{x+7}{x^2-x-6}\\ \frac{x+7}{(x-3)(x+2)}& =\frac{x+7}{(x-3)(x+2)}\\ & =\frac{A}{x+2}+\frac{B}{x-3}\\ & =\frac{A(x-3)}{(x+2)(x-3)}+\frac{B(x+2)}{(x+2)(x-3)}\end{array}$$

Now that we have isolated the factors with simple algebraic manipulation, we can try to solve for A and B.

$$\begin{array}{rl}\frac{x+7}{(x+2)(x-3)}& =\frac{A(x-3)}{(x+2)(x-3)}+\frac{B(x+2)}{(x+2)(x-3)}\\ x+7& =A(x-3)+B(x+2)\end{array}$$

We can expand that into:

$$x+7=Ax-3A+Bx+2B$$

And turn that into:

$$\begin{array}{rl}x+7& =(A+B)x-3A+2B\end{array}$$

Where we can set $A+B$ equal to the coefficient of x, or 1, and $-3A+B$ to 7, then solve through:

$$\begin{array}{r}A+B=1\\ -3A+2B=7\end{array}$$

$$\begin{array}{rl}A& =1-B\\ -3(1-B)+2B& =7\\ -3+3B+2B& =7\\ 5B& =10\\ B& =2\\ & \\ A+(2)& =1\\ A& =-1\end{array}$$

Now, we can write our partial fractional decomposition out with properly substituted values; recall the third line step of this whole process.

$$\begin{array}{rl}f(x)& =\frac{A}{x+2}+\frac{B}{x-3}\\ & =-\frac{-1}{x+2}+\frac{2}{x-3}\end{array}$$

Since we’re finding the antiderivative, it’s time to integrate f(x).

$$\int \frac{-1}{x+2}+\frac{2}{x-3}dx$$

Recall that the antiderivative of $\frac{1}{x}$ is $\ln x$; this is our essential concept for both these integrations. We can even rewrite it to look more familiar, then antiderivate:

$$\begin{array}{rl}\int f(x)dx& =\int -\frac{1}{x+2}+2\frac{1}{x-3}dx\\ & =-\ln |x+2|+2\ln |x-3|+C\end{array}$$

So our final antiderivative is:

$$-1(\ln |x+2|)+2(\ln |x-3|)+C$$

(As a note, $C$ is still $C$ because it is the combined constant of integration; $C_1+C_2=C$).

More PFD §

Let’s consider a few more problems.

$$\int \frac{19-2x}{x^2+x-6}dx$$

This problem is pretty identifiable as needing PFD; we can start by factoring the denominator into two parts:

$$\int \frac{19-2x}{(x-2)(x+3)}dx$$

Then we can proceed with splitting it up into fractions, then creating our system of equations

$$\begin{array}{rl}\frac{19-2x}{(x-2)(x+3)}& =\frac{A}{(x-2)}+\frac{B}{(x+3)}\\ 19-2x& =A(x+3)+B(x-2)\\ & \end{array}$$

$$\begin{array}{rl}A+B& =-2\\ 3A-2B& =19\\ B& =-A-2\\ & \\ 3A+2A+4& =19\\ 5A& =15\\ A& =3\\ & \\ B& =-3-2\\ B& =-5\end{array}$$

Then, we can substitute it back into the fractional equation and integrate term by term.

$$\begin{array}{rl}\frac{19-2x}{(x-2)(x+3)}& =\frac{3}{(x-2)}-\frac{5}{(x+3)}\\ \int \frac{3}{x-2}dx-\int \frac{5}{x+3}dx& =3\ln |x-2|-5\ln |x+3|+C\end{array}$$

Another example can be this:

$$\int \frac{3x+1}{2x^2+3x-2}dx$$

Integration by Parts §

Integration by parts is when you integrate. By parts. The equation is:

$$\int udv=uv-\int vdu$$

You essentially just need to know how to decide what u and v should be. We can look at it through this function:

$$\int x\times e^{5x}dx$$

We will choose based on what is easier to integrate, which is $e^{5x}$; that will be our v. Our u will then be $x$. We should find $du$ and $dv$ at the start.

$$\begin{array}{rlrl}u& =x& du& =dx\\ v& =\frac{1}{5}{e}^{5x}& dv& =e^{5x}dx\end{array}$$

And we can substitute into the initial equation:

$$\begin{array}{rl}\int udv& =uv-\int vdu\\ \int x{e}^{5x}dx& =\left(x\times \frac{1}{5}{e}^{5x}-\int \frac{1}{5}{e}^{5x}dx\right)\\ & =\frac{x{e}^{5x}}{5}-\frac{e^{5x}}{25}+C\end{array}$$