Indefinite integrals

What you’ll learn

• Difference between definite and indefinite integrals
• Key integration rules

Integration can be performed without or with boundaries.

1. Definite integrals (${\int }_a^{b}f(x)dx$):

Bound by an interval $[a,b]$, these calculate the net accumulated area under a curve and evaluate to a single number.

2. Indefinite integrals ($\int f(x)dx$):

These have no limits of integration. They reverse differentiation to find the general antiderivative, resulting in a family of functions that differ only by a constant.

The constant of integration ($+C$)

When finding an indefinite integral, you are reversing differentiation to find the original function. However, functions that differ only by a vertical shift share the exact same derivative. For example, consider the three functions:

$$\begin{array}{rl}F(x)& =x^2-1\\ G(x)& =x^2+100\\ H(x)& =x^2+\pi \end{array}$$

Because the derivative of any constant is $0$, differentiating all three yields the same result:

$$F^{\prime }(x)=G^{\prime }(x)=H^{\prime }(x)=2x$$

When integrating $2x$, we don’t know which specific vertical shift the original function had without additional information.

To account for this, indefinite integrals always include the constant of integration $+C$. The result of this indefinite integral

$$\int 2xdx=x^2+C$$

represents the entire family of functions whose derivative is $2x$.

Sidenote

Constant of integration

Your answer for an indefinite integral must include $+C$ at the end unless an initial condition (a specific point on the curve) is given to solve for $C$.

The reverse power rule handles the integration of most power functions:

$$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$$

for $n\ne -1$.

The special case of $n=-1$

If we attempt to integrate $f(x)=x^{-1}$ using the reverse power rule, we run into a division-by-zero problem:

$$\int \frac{1}{x}dx=\int x^{-1}dx=\frac{x^{-1+1}}{-1+1}=\frac{x^0}{0}$$

Instead, reverse a known derivative rule. Since

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

then

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

By working backward from derivative rules, we obtain a set of core integration rules:

Power and exponential:

$f(x)$	Indefinite integral
$x^n$	$\frac{x^{n+1}}{n+1}+C$
$\frac{1}{x}$	$\ln |x|+C$
$e^x$	$e^x+C$
$a^x$	$\frac{a^x}{\ln (a)}+C$

Trigonometric:

$f(x)$	Indefinite integral
$\sin x$	$-\cos x+C$
$\cos x$	$\sin x+C$
${\sec }^{2}x$	$\tan x+C$
${\csc }^{2}x$	$-\cot x+C$
$\sec x\tan x$	$\sec x+C$
$\csc x\cot x$	$-\csc x+C$

Inverse trigonometric:

$f(x)$	Indefinite integral
$\frac{1}{\sqrt{1-x^2}}$	${\sin }^{-1}(x)+C$
$\frac{1}{1+x^2}$	${\tan }^{-1}(x)+C$
$\frac{1}{|x|\sqrt{x^2-1}}$	${\sec }^{-1}(x)+C$

Note: The integrals for the co-functions (${\cos }^{-1}(x)$, ${\cot }^{-1}(x)$, and ${\csc }^{-1}(x)$) are simply the negative counterparts of the rules above, as established in section 3.6.

Example 1: Applying foundational rules

Find the indefinite integral

$$\int (3e^x+\frac{4}{x}-5\csc x\cot x)dx$$

Solution

Example 2: Rewriting before integrating

Find the indefinite integral

$$\int \frac{1-2\sin x}{{\cos }^{2}x}dx$$

Solution

Example 3: Finding $f(x)$

So far, our indefinite integrals have included $+C$ to represent the entire family of curves, known as the general solution.

However, if the problem gives an initial condition, which is a specific point that the original function passes through, we can anchor the graph and solve for an exact value of $C$ to one specific, or particular, solution to define the function.

If $f^{\prime }(x)=\frac{(x-2{)}^2}{2x}$ and $f(1)=5$, find the particular solution of $f(x)$.

Solution

Example 4: Using your calculator

In the previous section, we used the integral tool in Desmos to simplify the arithmetic involved in applying the FTC.

Similarly, some complicated integrals on the calculator-permitted sections of the exam are not intended to be evaluated by hand. You’ll save valuable time by using your calculator to evaluate the definite integral directly.

Sidenote

Desmos: Definite integral tool

Desmos requires you enter the bounds of integration - it won’t be able to give you the antiderivative itself.

A factory releases a pollutant into a river at a rate of $R(t)=\frac{500\sqrt{t}}{1+e^{-0.2t}}$ grams per hour, where time $t$ is the number of hours since the start of the day. To the nearest gram, how much pollutant was released between $5\le t\le 12$?

Solution