Indefinite Integrals of Polynomials

We are now going to look at a technique for finding the indefinite integrals of the simplest type of functions - polynomials!

Theorem 1: If $f$ is in the form $f(x) = bx^n$ where $b$ and $C$ are both constants, then $\int bx^n \: dx = \frac{bx^{n+1}}{n + 1} + C$.

• Proof: From the fundamental theorem of calculus part 1, we get that $\frac{d}{dx} \int f(x) \: dx = f(x)$. If we can differentiate both sides of $\int bx^n \: dx = \frac{bx^{n+1}}{n + 1} + C$ and prove their equality, then we have proven the equality of the integral.

(1)

\begin{align} \int bx^n \: dx = \frac{bx^{n+1}}{n + 1} + C \\ \frac{d}{dx} \int bx^n \: dx = \frac{d}{dx} \left [ \frac{bx^{n+1}}{n + 1} + C \right ] \\ bx^n = b (n+1) \frac{x^n}{n+1} \quad [\mathrm{By\:FTC1}] \\ bx^n = bx^n \quad \blacksquare \end{align}

We will now apply this important property to some examples.

Example 1

Evaluate the following indefinite integral $\int 5x^9 \: dx$.

Applying the rule from above, we end up obtaining that:

(2)

\begin{align} \int 5x^9 \: dx \\ = \frac{5x^{9+1}}{9 + 1} \\ = \frac{5x^{10}}{10} \\ = \frac{x^{10}}{2} \end{align}

Example 2

Evaluate the following indefinite integral $\int 5x^2(3x - 9x^4) \: dx$.

We must first simplify this integral before evaluating it:

(3)

\begin{align} \int 5x^2(3x - 9x^4) \: dx \\ = \int 15x^3 - 45x^6 \: dx \\ = \int 15x^3 \: dx - \int 45x^6 \: dx \\ = \frac{15x^{3+1}}{3 + 1} - \frac{45x^{6+1}}{6 + 1} \\ = \frac{15x^{4}}{4} - \frac{45x^{7}}{7} \\ \end{align}

Example 3

Evaluate the following indefinite integral $\int (-2x^8)^2 -2x^8 \: dx$.

Once again, we must first simplify this integral before evaluating it:

(4)

\begin{align} \int (-2x^8)^2 -2x^8 \: dx \\ = \int 4x^{16} -2x^8 \: dx \\ = \int 4x^{16} \: dx - \int 2x^8 \: dx\\ = \frac{4x^{16+1}}{16+1} - \frac{2x^{8+1}}{8+1} \: dx \\ = \frac{4x^{17}}{17} - \frac{2x^{9}}{9} \: dx \\ \end{align}