8  Integrals

Integration is one of the two central operations in calculus, alongside differentiation. While differentiation measures instantaneous change, integration measures accumulated quantity. In simple terms, an integral adds up infinitely many tiny pieces to form a total [1].

8.1 Illustrations for Integrals

Imagine dividing a quantity into extremely small parts—lengths, areas, volumes, masses, or any measurable quantities. An integral is the limit of the sum of these tiny parts [2]. The integral symbol \(\int\) was introduced by Leibniz and resembles an elongated “S”, meaning “sum” [3].

Components:

• \(\int\) : integral (sum) symbol
  
• \(f(x)\) : function being integrated
  
• \(dx\) : infinitesimally small change in \(x\) [4]
• \(a, b\) : lower and upper bounds (for definite integrals)

Example:

\[ \int_0^3 x.dx \]

General formula:

\[ \int x^{n}\, dx = \frac{x^{n+1}}{n+1} + C \]

Apply for ( n = 1 ):

\[ \int x\, dx = \frac{x^{2}}{2} \]

Evaluate the definite integral:

\[ \int_{0}^{3} x\, dx = \left[ \frac{x^2}{2} \right]_{0}^{3} \] \[ = \frac{3^2}{2} - \frac{0^2}{2} \] \[ = \frac{9}{2} = 4.5 \]

In General a function \(f(x)\), we approximate the area under the curve using rectangles of width \(\Delta x\):

\[ \text{Approximate Area} \approx \sum_{i=1}^{n} f(x_i)\Delta x \]

As the rectangles become thinner (\(\Delta x \to 0\)), the approximation becomes exact:

\[ \int_a^b f(x)\, dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x \]

This is called a definite integral.

8.2 Two Types of Integrals

8.2.1 Indefinite Integral

An indefinite integral represents a family of functions whose derivative is \(f(x)\) [1] [2]:

\[ \int f(x)\, dx = F(x) + C \]

where \(F'(x) = f(x)\) and \(C\) is the constant of integration. This is the reverse process of differentiation [4]. Therefore, the most important link between derivatives and integrals is:

\[ \frac{d}{dx}\left(\int f(t)\, dt\right) = f(x) \]

8.2.2 Definite Integral

A definite integral computes the actual numerical value of accumulated area or quantity between \(x=a\) and \(x=b\) [5]:

\[ \int_a^b f(x)\, dx = F(b) - F(a) \]

It has no “\(+C\)” because the value is a number, not a function [6].

8.3 Area in Single Function

When we want to compute the area bounded by a single function \(y = f(x)\) over a certain interval, we use the definite integral [7]. This is one of the most important ideas in calculus because it allows us to measure areas under curves that are not simple geometric shapes [8].

8.3.1 Parabola

Compute the area under:

\[ f(x) = x^2 \]

on the interval \(0 \le x \le 2\).

\[ A = \int_0^2 x^2\, dx \]

Integrating:

\[ \int x^2\, dx = \frac{x^3}{3} \]

Substitute the bounds:

\[ A = \left[ \frac{x^3}{3} \right]_0^2 = \frac{8}{3} \]

Final Answer:
\[A = \frac{8}{3}\]

8.3.2 Linear Function

Find the area under:

\[ f(x) = -x + 4 \]

from \(x=0\) to \(x=4\).

\[ A = \int_0^4 (-x + 4)\, dx \]

Integrate:

\[ \int (-x + 4)\, dx = -\frac{x^2}{2} + 4x \]

Substitute:

\[ A = \left[-\frac{x^2}{2} + 4x\right]_0^4 = 8 \]

Final Answer:
\[A = 8\]

8.3.3 Function Crossing the x-axis

Consider:

\[ f(x) = x - 2 \]

on \([0, 4]\).
The function changes sign at \(x=2\).

Thus:

\[ A = \int_0^2 |x - 2|\, dx + \int_2^4 |x - 2|\, dx \]

For \(0 \le x < 2\):

\[ |x - 2| = 2 - x \]

For \(2 \le x \le 4\):

\[ |x - 2| = x - 2 \]

Compute each:

\[ A_1 = \int_0^2 (2 - x)\, dx = 2 \]

\[ A_2 = \int_2^4 (x - 2)\, dx = 2 \]

Total area:

\[ A = 2 + 2 = 4 \]

Final Answer:
\[A = 4\]

8.3.4 Visual Explanation (Video 3)

The following video helps visualize area under a curve:

8.4 Area in Two Function

When finding the area between two curves, we no longer measure the area under a single function. Instead, we measure the vertical distance between two functions across an interval.

8.4.1 Concept

Suppose we have two functions:

• \(f(x)\) — the upper curve
  
• \(g(x)\) — the lower curve

on the interval \([a, b]\). The area between them is the accumulated difference in height:

\[ \text{Area} = \int_{a}^{b} \left( f(x) - g(x) \right)\, dx \]

The integrand is always upper minus lower, ensuring the area remains positive.

8.4.2 Steps to Compute the Area

Problem: Find the area between the curves \(y = x\) and \(y = x^2\) from \(x = 0\) to \(x = 1\).

1. Identify which function is on top
   For \(0 \le x \le 1\) we have \(x \ge x^2\).
   Thus \(f(x)=x\) is the upper curve and \(g(x)=x^2\) is the lower curve.

2. Find the intersection points
   Solve \(f(x)=g(x)\): \[ x = x^2 \quad\Longrightarrow\quad x(1-x)=0 \quad\Longrightarrow\quad x=0,\; x=1. \] These are the bounds of the region.

3. Set up the integral
   The area is \[ \text{Area}=\int_{0}^{1}\big(f(x)-g(x)\big)\,dx=\int_{0}^{1}\big(x-x^2\big)\,dx. \]

4. Evaluate the definite integral
   Find an antiderivative: \[ \int (x-x^2)\,dx=\frac{x^2}{2}-\frac{x^3}{3}+C. \] Evaluate from 0 to 1: \[ \begin{aligned} \text{Area} &=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{1} \\[4pt] &=\left(\frac{1^2}{2}-\frac{1^3}{3}\right)-\left(\frac{0^2}{2}-\frac{0^3}{3}\right) \\[4pt] &=\frac{1}{2}-\frac{1}{3} \\[4pt] &=\frac{3}{6}-\frac{2}{6} \\[4pt] &=\frac{1}{6}. \end{aligned} \]

Final answer:
\[\boxed{\text{Area} = \dfrac{1}{6}}\]

8.4.3 Visual Explanation (Video 2)

8.5 Volume using Integral

A solid of revolution is formed when a region in the plane is rotated around a line (axis of rotation), such as [1] [2]:

• the \(x\)-axis
  
• the \(y\)-axis
  
• any vertical or horizontal line

To compute its volume, we integrate the volume of infinitesimally thin slices [4]. In calculus, we compute the volume of a solid using definite integrals. Common methods include: Disk Method, Washer Method, and Shell Method [6] [5].

8.5.1 Disk Method

Used when the region touches the axis of rotation (no hole in the middle). If a region bounded by \(y = f(x)\) is rotated around the \(x\)-axis, the volume is:

\(V = \int_{a}^{b} \pi (f(x))^{2} \, dx\)

Each slice is a disk with:

• radius = \(f(x)\)
  
• thickness = \(dx\)

Let consider the following visualization, \(f(x)=x^2\)

8.5.2 Washer Method

Used when there is a gap between the region and the axis of rotation, producing a “hole”. If the outer radius is \(R(x)\) and the inner radius is \(r(x)\), the volume is:

\(V = \int_{a}^{b} \pi \left[ R(x)^{2} - r(x)^{2} \right] \, dx\)

This represents: Outer disk area minus inner disk area

8.5.3 Shell Method

Useful when rotating around the \(y\)-axis or when the function is easier to express in terms of \(x\). If the region is rotated around the \(y\)-axis, the volume is:

\(V = \int_{a}^{b} 2\pi x \, f(x) \, dx\)

Each slice is a hollow cylindrical shell with:

• radius = \(x\)
  
• height = \(f(x)\)
  
• thickness = \(dx\)

8.6 Example Problem

Find the volume of the solid obtained by rotating the region bounded by
\(y = x^2\) and \(y = 0\), for \(0 \le x \le 1\), around the x-axis.

Solution:

Because we revolve around the x-axis, we use the Disk Method.

• Volume formula using disk method: \(V = \pi \int_{0}^{1} (x^2)^2 \, dx\)
• Simplify the integrand: \(V = \pi \int_{0}^{1} x^4 \, dx\)
• Evaluate the integral: \(V = \pi \left[ \frac{x^5}{5} \right]_{0}^{1}\)
• Final answer: \(V = \frac{\pi}{5}\)

Therefore, the volume of the solid is: \(\frac{\pi}{5}\)

8.6.1 Visual Explanation (Video 3)

References

[1]

Stewart, J., Calculus: Early transcendentals, Cengage Learning, 2016

[2]

Thomas, G. B., Weir, M. D., and Hass, J., Thomas’ calculus, Pearson, 2014

[3]

Leibniz, G. W., Historia et origo calculi differentialis, 1693

[4]

Apostol, T. M., Calculus, vol. 1: One-variable calculus with an introduction to linear algebra, Wiley, 1967

[5]

Anton, H., Bivens, I., and Davis, S., Calculus, Wiley, 2013

[6]

Larson, R. and Edwards, B., Calculus, Brooks/Cole, 2013

[7]

Spivak, M., Calculus, Publish or Perish, 2008

[8]

Courant, R. and John, F., Introduction to calculus and analysis, vol. 1, Springer, 1999