Integrals of constants and powers

We know that integration is the inverse operation to differentiation. Furthermore, we know we can use the Fundamental Theorem of Calculus to find the exact value of a definite integral, by evaluating the antiderivative of the integrand function at the limits of integration. However, we still need to learn how to determine these antiderivatives. In this topic, we'll get acquainted with the rules to determine the antiderivative of a few simple but very common functions.

Integral of a constant

We know that integration is a linear operation. Therefore, just like with derivatives, we can take the constant factor out of the integral:

$$\int \text{k} \cdot f(x) dx = \text{k} \int f(x) dx$$We also know that the antiderivative of $f(x)=1$ is $x + C$, since

$$\frac{d}{dx}F(x) = 1 \quad \text{with} \quad F(x) = x + C$$Then,

$$\int \text{k} dx = \text{k} \int 1 dx = \text{k} \ (x+C) = \text{k}x + kC = \text{k}x + C' \quad ; \quad C' = kC$$Since $C$ is an arbitrary constant, $C'$ would be just another arbitrary constant. Because of this, it is not unusual to find this rule as:

$$\int \text{k} dx = \text{k}x + C$$For example,

$$\begin{align*} &\int 2 dx = 2 \cdot \int dx = 2x + C \\ \\ &\int -5 dx = -5 \cdot \int dx = -5x + C \end{align*}$$

Integral of a power

In order to determine the integral, let's recall what the derivative of a power function looks like:

$$\frac{d}{dx}(x^m) = m \cdot x^{m-1}$$Now, let's define: $n=m-1$

$$\frac{d}{dx}(x^{n+1}) = (n+1) \cdot x^n$$ Solving for $x^n$

$$x^n =\frac{\frac{d}{dx}(x^{n+1})}{(n+1)}$$Integrating both sides,

$$\int x^n dx = \int \frac{\frac{d}{dx}(x^{n+1})}{(n+1)} dx = \frac{1}{n+1} \int \frac{d}{dx}(x^{n+1}) dx$$Inverse operations cancel out (but we still need our constant of integration since we are computing a general antiderivative!

$$\frac{1}{n+1} \int \frac{d}{dx}(x^{n+1}) dx = \frac{1}{n+1} \cdot x^{n+1} + C$$Then,

$$\int \text{k}\cdot x^n dx = \text{k} \cdot \frac{x^{n+1}}{n+1} + C \quad ; \ n \neq -1$$As you can see, the antiderivative becomes undefined when $n=-1$

For example,

$$\begin{align*} &\int x^2 dx = \frac{x^{2+1}}{2 + 1} + C = \frac{x^3}{3} + C \\ \\ &\int x^{-5} dx = \frac{x^{-5+1}}{-5 + 1} + C = -\frac{x^{-4}}{4} + C \\ \\ &\int x^{1/2} dx = \frac{x^{\frac{1}{2}+1}}{\frac{1}{2} + 1} + C = \frac{2}{3} x^{3/2} + C \end{align*}$$

Integral of a polynomial

We can bring together the rules we have so far to figure out the indefinite integral of a polynomial of the form:

$$p(x) = a_n \cdot x^n + a_{n-1} \cdot x^{n-1} + \cdots + a_2 \cdot x^2 + a_1 \cdot x + a_0$$where $a_i$ is a constant coefficient for each term.

Since integration is a linear operation, the integral of a sum is the sum of the integrals of each summand:

$$\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx$$Then,

$$\begin{align*} \int p(x) dx &= \int a_n \cdot x^n + a_{n-1} \cdot x^{n-1} + \cdots + a_2 \cdot x^2 + a_1 \cdot x + a_0 \ dx \\ &= \int a_n \cdot x^n \ dx + \int a_{n-1} \cdot x^{n-1} \ dx + \cdots + \int a_2 \cdot x^2 \ dx + \int a_1 \cdot x \ dx + \int a_0 \ dx \\ \int p(x) dx &= a_n \cdot \frac{x^{n+1}}{n+1} + a_{n-1} \cdot \frac{x^n}{n} + \cdots + a_2 \cdot \frac{x^3}{3} + a_1 \cdot \frac{x^2}{2} + a_0 \cdot x + C \\ \end{align*}$$For example,

$$\begin{align*} \int (5x^2 + x -2) dx &= \int 5x^2 dx + \int x dx + \int (-2) dx \\ & = 5\int x^2 dx + \int x dx -2 \int dx \\ & = \frac{5x^3}{3} + \frac{x^2}{2}-2x+ C = \frac{1}{6} (10x^3+3x^2-12x) + C \end{align*}$$

Integral of reciprocal of x

As stated above, the antiderivative of a power function $\text{k} \cdot x^n$ becomes undefined for $n=-1$. So, what do we do in this case? Fear not, $f(x)=\frac{1}{x}$ has an antiderivative of its own!

We already know that

$$\frac{d}{dx}\ln(x) = \frac{1}{x}$$Then,

$$\int \frac{1}{x}dx=\ln(x) + C$$However, this does not work for $x \leq 0$ , which is why we define this antiderivative as

$$\int \frac{1}{x}dx=\ln|x| + C$$Since

$$\frac{d}{dx}\ln|x| = \frac{1}{x}$$as well!

Conclusion

In this topic, we've learned the rules to determine antiderivatives for commonly used functions such as:

• Constant functions: $$f(x)=k \implies \int f(x) dx = \text{k}x + C$$
• Power functions: $$f(x)=k \cdot x^n \implies \int f(x) dx = \begin{cases} \text{k} \cdot \frac{x^{n+1}}{n+1} + C \quad &; \ n \neq -1 \\ \ln(x) + C \quad & ; \ n = -1 \end{cases}$$
• Polynomials: $$f(x) = a_n \cdot x^n + \cdots +a_1 \cdot x + a_0 \implies \int f(x) dx = a_n \cdot \frac{x^{n+1}}{n+1} + a_{n-1} \cdot \frac{x^n}{n} + \cdots + a_2 \cdot \frac{x^3}{3} + a_1 \cdot \frac{x^2}{2} + a_0 \cdot x + C$$