Integration

Beginning Integration — The Reverse of Differentiation

Introduction

Integration and differentiation are the two most important operations in calculus. Specifically, integration is the reverse process of differentiation. To integrate is to find a function whose derivative is given. Thus, given a differential equation, \(F'(x) = \frac{dF}{dx} = \frac{dy}{dx} = f(x)\),

where:

The parts of the indefinite integral

\(y = F(x)\)	the anti-derivative or the solution of the differential equation
\(f(x)\,dx\)	the integrand of the integral
\(c\)	the constant of integration or arbitrary constant
\(\int\)	the integral sign

The equation \(\int f(x)\,dx = F(x) + c\) is called an Indefinite Integral.

The following are some of the basic integration formulas:

Formula I

If \(f(x) = \frac{dy}{dx} = x^n\), where \(n\) is a real number not equal to \(-1\), then:

To show differentiation, \(\frac{d}{dx}\left(\frac{1}{n+1}x^{n+1}\right) = x^{n}\).

Example

a. Calculate \(\int x^{7}\,dx\)

Solution

1. Step 1 — Apply the rule

   Use \(\int x^{n}\,dx = \frac{1}{n+1}x^{n+1} + c\) and substitute \(n = 7\):

   \[ \int x^{7}\,dx = \frac{1}{7+1}x^{7+1} + c \]
2. Step 2 — Simplify
   \[ \int x^{7}\,dx = \frac{1}{8}x^{8} + c \]

Example

b. Calculate \(\int \frac{1}{\sqrt[3]{x}}\,dx\)

Solution

1. Step 1 — Express in index form

   Expressing \(\frac{1}{\sqrt[3]{x}}\) in the form \(x^{n}\):

   \[ \int \frac{1}{\sqrt[3]{x}}\,dx = \int x^{-\frac{1}{3}}\,dx \]
2. Step 2 — Apply the rule

   Use \(\int x^{n}\,dx = \frac{1}{n+1}x^{n+1} + c\) and substitute \(n = -\frac{1}{3}\):

   \[ = \frac{1}{-\frac{1}{3}+1}\,x^{-\frac{1}{3}+1} + c \]
3. Step 3 — Add the indices

   Add: \(-\frac{1}{3} + 1 = \frac{-1+3}{3} = \frac{2}{3}\):

   \[ = \frac{x^{\frac{2}{3}}}{\frac{2}{3}} + c \]
4. Step 4 — Simplify the fraction

   Divide \(x^{\frac{2}{3}}\) by \(\frac{2}{3}\):

   \[ = \frac{3x^{\frac{2}{3}}}{2} + c \]
5. Step 5 — Express in radical form

   Expressing in radical form:

   \[ \int \frac{1}{\sqrt[3]{x}}\,dx = \frac{3\sqrt[3]{x^{2}}}{2} + c \]

The condition \(n \neq -1\) exists because the rule would require dividing by zero when \(n = -1\). The integral of \(x^{-1}\) has its own result:

where \(\ln\) is the natural logarithm and the modulus ensures the result is defined for negative values of \(x\) as well. You will meet this integral properly when you study logarithms; for now, note only that the power rule is never used when \(n = -1\).

Formula II — Integral of a function times a constant

If \(f(x)\) is an integrable function and \(k\) is any constant, then:

Following this general formula, if \(f(x) = \frac{dy}{dx} = ax^{n}\), where \(a\) is any constant, then:

Example

Calculate \(\int 5x^{4}\,dx\)

Solution

1. Step 1 — Write the constant, 5, before the integral sign
   \[ \int 5x^{4}\,dx = 5\int x^{4}\,dx \]
2. Step 2 — Apply the rule

   Use \(a\int x^{n}\,dx = \frac{ax^{n+1}}{n+1} + c\), substitute \(n = 4\):

   \[ = \left(\frac{5x^{4+1}}{4+1}\right) + c \]
3. Step 3 — Evaluate the index and denominator

   \(n + 1 = 4 + 1 = 5\):

   \[ = \left(\frac{5x^{5}}{5}\right) + c \]
4. Step 4 — Simplify

   Dividing 5 by 5:

   \[ \int 5x^{4}\,dx = x^{5} + c \]

Formula III — Integrating a Sum

If \(f(x)\) and \(g(x)\) are integrable functions, then:

In integrating a sum, each term is integrated separately. This formula can be extended to more than two functions.

Example

Calculate \(\int (6x^{6} + 4x^{\frac{2}{3}} + 8)\,dx\)

Solution

1. Step 1 — Split the integral term by term
   \[ \int (6x^{6} + 4x^{\frac{2}{3}} + 8)\,dx \]

   Use \(\int f(x)\,dx + g(x)\,dx = \int f(x)\,dx + \int g(x)\,dx\):

   \[ = \int 6x^{6}\,dx + \int 4x^{\frac{2}{3}}\,dx + \int 8\,dx \]
2. Step 2 — Apply the rule to each term

   Use \(\int ax^{n}\,dx = a\int x^{n}\,dx = \frac{ax^{n+1}}{n+1} + c\):

   \[ = \left(\frac{6x^{6+1}}{6+1}\right) + \left(\frac{4x^{\frac{2}{3}+1}}{\frac{2}{3}+1}\right) + 8x \]
3. Step 3 — Evaluate the indices
   \[ = \frac{6x^{7}}{7} + \left(\frac{4x^{\frac{5}{3}}}{\frac{5}{3}}\right) + 8x \]
4. Step 4 — Simplify the second term

   Divide 4 by \(\frac{5}{3}\) of the second term to simplify:

   \[ \int (6x^{6} + 4x^{\frac{2}{3}} + 8)\,dx = \frac{6x^{7}}{7} + \frac{12x^{\frac{5}{3}}}{5} + 8x \]

Formula IV

Given a higher order derivative \(\frac{d^{2}y}{dx^{2}}\), then:

Example

If \(\frac{d^{2}y}{dx^{2}} = 10 - 3x\), find \(\frac{dy}{dx} + c\)

Solution

1. Step 1 — Split the integral term by term

   Use \(\int f(x)\,dx + g(x)\,dx = \int f(x)\,dx + \int g(x)\,dx\):

   \[ \frac{dy}{dx} = \int (10 - 3x)\,dx = \int 10\,dx - \int 3x\,dx \]
2. Step 2 — Apply the rule

   Use \(\int ax^{n}\,dx = a\int x^{n}\,dx = \frac{ax^{n+1}}{n+1} + c\):

   \[ = 10x - \left(\frac{3x^{1+1}}{1+1}\right) + c \]
3. Step 3 — Simplify
   \[ = 10x - \left(\frac{3x^{2}}{2}\right) + c \]
4. Step 4 — State the result
   \[ \frac{dy}{dx} = 10x - \frac{3}{2}x^{2} + c \]