MH1810 Math

Chapter 6a: Integration

Imron Rosyadi

Chapter 6: Integration

Welcome to Chapter 6! Today, we’ll dive into the fascinating world of integration, the inverse operation of differentiation. We’ll start with antiderivatives and indefinite integrals, then move on to definite integrals and their connection to areas. We’ll also explore various techniques for integration and touch upon improper integrals and numerical methods.

6.1 Antiderivatives & Indefinite Integral

We’ve mastered differentiation, finding the rate of change of a function. Now, let’s reverse that process. Given a function \(f(x)\), can we find another function \(F(x)\) such that \(F'(x) = f(x)\)?

If such an \(F(x)\) exists, it’s called an antiderivative of \(f(x)\).

The process of finding \(F(x)\) is called integration.

Note

Definition 6.1.1: Antiderivative

A function \(F\) is an antiderivative of \(f\) on an interval \((a, b)\) if \(F'(x) = f(x)\) for all \(x\) in \((a, b)\).

Think of it like this: if differentiation tells you the speed of a car, integration helps you figure out its position given its speed. It’s a fundamental concept in calculus.

Antiderivatives: Examples

Let’s look at some examples to solidify the concept:

Example (a):

We know \(\frac{d}{dx}(\sin x) = \cos x\) on \(\mathbb{R}\).

This means \(\cos x\) is the derivative of \(\sin x\).

Therefore, \(\sin x\) is an antiderivative of \(\cos x\).

Example (b):

Consider \(\frac{d}{dx}\left(x^3 - 4\sqrt{x} + 179\right)\).

Applying differentiation rules:

\(\frac{d}{dx}\left(x^3\right) = 3x^2\)

\(\frac{d}{dx}\left(-4\sqrt{x}\right) = \frac{d}{dx}\left(-4x^{1/2}\right) = -4 \cdot \frac{1}{2} x^{-1/2} = -2x^{-1/2} = -\frac{2}{\sqrt{x}}\)

\(\frac{d}{dx}\left(179\right) = 0\)

So, \(\frac{d}{dx}\left(x^3 - 4\sqrt{x} + 179\right) = 3x^2 - \frac{2}{\sqrt{x}}\) on \((0, \infty)\).

This means \(\left(x^{3} - 4\sqrt{x} +179\right)\) is an antiderivative of \(3x^{2} - \frac{2}{\sqrt{x}}\).

Notice how the constant term, like 179 in example (b), vanishes when we differentiate. This leads us to an important property of antiderivatives.

General Antiderivatives

What if we had \(x^3 - 4\sqrt{x} + 5\) instead of \(x^3 - 4\sqrt{x} + 179\)?

Its derivative would still be \(3x^2 - \frac{2}{\sqrt{x}}\).

This illustrates a crucial point: if \(F(x)\) is an antiderivative of \(f(x)\), then \(F(x) + C\) (where \(C\) is any constant) is also an antiderivative.

Important

Theorem 6.1.3: General Antiderivative

If \(F\) is an antiderivative of \(f\) on an interval \(I\), then the most general antiderivative of \(f\) on \(I\) is \(F(x) + C\), where \(C\) is any arbitrary constant.

Example 6.1.4:

Since \(\frac{d}{dx} (\sin x) = \cos x\), then \(\sin x\) is an antiderivative of \(\cos x\).

The most general antiderivative of \(\cos x\) is \(\sin x + C\).

This arbitrary constant ‘C’ is often called the “constant of integration.” It represents all possible vertical shifts of the antiderivative function that would still have the same derivative.

Indefinite Integrals

We have a special notation for the most general antiderivative.

Note

Definition 6.1.5: Indefinite Integral

The indefinite integral of \(f\), denoted by \(\int f(x) dx\), is the most general antiderivative of \(f\).

The function \(f\) is called the integrand.

The process of finding an antiderivative or an indefinite integral is called integration. \[ \int f(x) dx \]

Remark: By definition, we have: \[ \frac{d}{dx} \left(\int f(x) dx\right) = f(x) \]

Tip

This remark highlights the inverse relationship between differentiation and integration. If you integrate a function and then differentiate the result, you should get back your original function.

Indefinite Integrals: Examples

Let’s apply the definition:

Example 6.1.6: \(\int \cos x \, dx = \sin x + C\)

Example 6.1.7:

\(\int \left(3x^{2} - \frac{2}{\sqrt{x}}\right)dx = x^3 - 4\sqrt{x} + C\)

Example 6.1.8: Prove that \(\int \frac{1}{x} dx = \ln |x| + C\).

Solution:

It suffices for us to prove that the derivative of \(\ln |x|\) is \(\frac{1}{x}\).

We know \(\frac{d}{dx} (\ln x) = \frac{1}{x}\) for \(x > 0\).

For \(x < 0\), let \(y = \ln |x| = \ln(-x)\).

Using the chain rule, \(\frac{dy}{dx} = \frac{1}{-x} \cdot \frac{d}{dx}(-x) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}\).

Therefore, \(\frac{d}{dx} \ln |x| = \frac{1}{x}\).

Hence, \(\int \frac{1}{x} dx = \ln |x| + C\).

In Example 6.1.7, we just reversed the differentiation from Example 6.1.2. The constant ‘C’ is essential here. Example 6.1.8 is a good proof to remember.

6.1.1 Rules for Integration

Just like differentiation, integration has rules to make the process easier.

Important

Theorem 6.1.9 (Rules for integration):

1. \(\int (f(x) + g(x))dx = \int f(x)dx + \int g(x)dx.\)
2. \(\int (f(x) - g(x))dx = \int f(x)dx - \int g(x)dx.\)
3. \(\int cf(x)dx = c\int f(x)dx.\)

These rules are direct consequences of the corresponding rules for differentiation. The integral of a sum is the sum of the integrals, and constants can be pulled out of the integral. Very useful for breaking down complex integrals.

Integration Rules: Examples

Let’s practice using these rules.

Example 6.1.10:

\(\int \left(2x^3 + 3x^{\frac{3}{2}}\right) dx\)

\(= 2\int x^3 dx + 3\int x^{\frac{3}{2}} dx\)

\(= 2\left(\frac{x^{3+1}}{3+1}\right) + 3\left(\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right) + C\)

\(= 2\left(\frac{x^4}{4}\right) + 3\left(\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right) + C\)

\(= \frac{1}{2} x^4 + \frac{6}{5} x^{\frac{5}{2}} + C\)

Example 6.1.11:

\(\int (4u^{-5} - 2\cos u + e^u)du\)

\(= 4\int u^{-5}du - 2\int \cos u du + \int e^u du\)

\(= 4\left(\frac{u^{-4}}{-4}\right) - 2(\sin u) + e^u +C\)

\(= -u^{-4} - 2\sin u + e^u +C\)

Remember the power rule for integration: \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) for \(n \neq -1\). Also, recall basic antiderivatives like \(\int \cos x dx = \sin x + C\) and \(\int e^x dx = e^x + C\).

Integration Rules: More Examples

Sometimes, you need to simplify the integrand first.

Example 6.1.12:

\(\int \frac{(1 + x^2)^2}{x^4} dx\)

First, expand the numerator: \((1 + x^2)^2 = 1 + 2x^2 + x^4\).

Then divide by \(x^4\):

\(\int \left(\frac{1}{x^4} + \frac{2x^2}{x^4} + \frac{x^4}{x^4}\right)dx\)

\(= \int (x^{-4} + 2x^{-2} + 1)dx\)

Now integrate term by term:

\(= \frac{x^{-3}}{-3} + 2\frac{x^{-1}}{-1} + x + C\)

\(= \frac{-1}{3} x^{-3} - 2x^{-1} + x + C\)

Example 6.1.13:

\(\int \left(\frac{1}{\sqrt{t}} + \frac{\pi}{\sqrt{1 - t^2}}\right) dt\)

Rewrite \(\frac{1}{\sqrt{t}}\) as \(t^{-1/2}\):

\(= \int t^{-1/2} dt + \pi \int \frac{1}{\sqrt{1 - t^2}} dt\)

Recall that \(\frac{d}{dt}(\sin^{-1}t) = \frac{1}{\sqrt{1-t^2}}\).

\(= \frac{t^{1/2}}{1/2} + \pi \sin^{-1}(t) + C\)

\(= 2\sqrt{t} + \pi \sin^{-1}(t) + C\)

These examples show the importance of algebraic manipulation before integrating. Also, knowing your basic derivative and antiderivative pairs is crucial. The \(\frac{1}{\sqrt{1-t^2}}\) term directly links to inverse trigonometric functions.

Finding the Constant of Integration

Sometimes, we’re given an initial condition that helps us find the specific value of \(C\).

Example 6.1.14: If \(f'(x) = 2x - 3\) and \(f(2) = 3\), find \(f(x)\).

Solution:

1. Find the general antiderivative:

   If \(f'(x) = 2x - 3\), then \(f(x) = \int (2x - 3) \, dx\).

   \(f(x) = 2\int x \, dx - 3\int 1 \, dx\)

   \(f(x) = 2\left(\frac{x^2}{2}\right) - 3(x) + C\)

   \(f(x) = x^2 - 3x + C\) for some constant \(C\).

2. Use the initial condition to find \(C\):

   We are given \(f(2) = 3\). Substitute \(x=2\) and \(f(x)=3\) into the equation:

   \(3 = (2)^2 - 3(2) + C\)

   \(3 = 4 - 6 + C\)

   \(3 = -2 + C\)

   \(C = 5\)

3. State the specific function:

   Thus, \(f(x) = x^2 - 3x + 5\).

Tip

This is a common application of indefinite integrals: finding a particular function given its rate of change and a point it passes through. This is called solving an initial value problem.

Beyond Basic Rules

The integration rules we’ve covered are fundamental, but many integrals require more advanced techniques.

There are integration rules that correspond to the product rule and the chain rule for differentiation.

These will be discussed later in the chapter.

They lead to special integration methods, namely:

• Integration by Parts (corresponding to the product rule)
• Substitution Rule (corresponding to the chain rule)

We will also explore techniques for integrating rational functions using partial fractions.

Illustration of various integral symbols and mathematical functions, representing the complexity and breadth of integration techniques.

Just like differentiation had simple rules and then more complex ones like the product and chain rules, integration also has its specialized techniques. Integration by parts and substitution are incredibly powerful tools you’ll use constantly in higher-level math and engineering.

6.2 The Definite Integral and Area Under a Curve

Moving from indefinite to definite integrals, we connect integration to a very intuitive geometric concept: area.

Area Under a Curve \(y = f(x)\), \(f(x) > 0\)

To find the area under a curve \(y = f(x)\), where \(f(x) > 0\) from \(x = a\) to \(x = b\):

1. Divide the interval: We divide the interval \([a, b]\) into \(n\) equal subintervals:

   \(\left[ x _ {0}, x _ {1} \right], \left[ x _ {1}, x _ {2} \right], \dots , \left[ x _ {k - 1}, x _ {k} \right], \dots \left[ x _ {n - 1}, x _ {n} \right].\)

2. Width of subintervals: The width of each subinterval is \(\Delta x = x_{k} - x_{k - 1} = \frac{b - a}{n}\).

   We have \(x_0 = a\) and \(x_n = b\). So, \(x _ {k} = a + k \left(\frac {b - a}{n}\right)\) for \(k = 0, 1, 2, \ldots , n\).

This is the heart of how we approximate areas using rectangles. We’re essentially chopping up the area into thin vertical strips.

Approximating Area with Rectangles

3. Choose sample points: In each \(k^{th}\) subinterval \([x_{k-1}, x_k]\), we choose a point \(x_k^*\) and evaluate the function value \(f(x_k^*)\).

4. Area of \(k^{th}\) rectangle: The area of the \(k^{th}\) rectangle, over \([x_{k-1}, x_k]\), with height \(f(x_k^*)\), is:

   \(f (x _ {k} ^ {*}) \Delta x = \frac {b - a}{n} f (x _ {k} ^ {*}).\)

5. Approximate total area: We approximate the area under the curve \(y = f(x)\) by the total areas of all these rectangles:

   \(\sum_ {k = 1} ^ {n} \frac {b - a}{n} f (x _ {k} ^ {*}).\)

As we increase the number \(n\) of subintervals, the length of the subinterval \(\Delta x\) tends to zero.

The approximations should approach the true area \(A\) under the curve:

\[ A = \lim _ {n \to \infty} \sum_ {k = 1} ^ {n} \frac {b - a}{n} f (x _ {k} ^ {*}). \]

This summation is crucial. It’s called a Riemann sum. The choice of \(x_k^*\) (left endpoint, right endpoint, midpoint) affects the approximation, but as n approaches infinity, the limit is the same for continuous functions.

Riemann Sum

The approximation process we just described leads to the definition of a Riemann sum.

Note

Definition 6.2.1: Riemann Sum

Let \(f\) be a function on \([a, b]\) and \(x _ {k} = a + k \left(\frac {b - a}{n}\right)\) for \(k = 0, 1, 2, \ldots n\).

With \(x_{k}^{*}\in [x_{k - 1},x_{k}]\), the finite sum

\[ \sum_ {k = 1} ^ {n} \frac {b - a}{n} f (x _ {k} ^ {*}), \] is called a Riemann sum of \(f\) on \([a, b]\).

Visual representation of a Riemann sum using rectangles to approximate the area under a curve.

The image clearly shows how the rectangles approximate the area. The thinner the rectangles (larger n), the better the approximation.

Riemann Sum: Example

Let’s construct a Riemann sum for a specific function.

Example 6.2.2: Riemann sum of \(f(x) = x^2\) on \([1,3]\).

For \(k = 1,2,3\ldots ,n\), the subinterval endpoints are:

\(x _ {k} = 1 + k \left(\frac {3 - 1}{n}\right) = 1 + \frac {2 k}{n}\).

The width of each subinterval is \(\Delta x = \frac{b-a}{n} = \frac{3-1}{n} = \frac{2}{n}\).

Suppose we take \(x_{k}^{*} = x_{k}\), the right end point of the \(k\)th subinterval.

We have the following Riemann sum for \(f(x)\) on \([1,3]\): \[ \sum_ {k = 1} ^ {n} \frac {2}{n} f (x _ {k} ^ {*}) = \sum_ {k = 1} ^ {n} (\frac {2}{n}) \left(1 + \frac {2 k}{n}\right) ^ {2}. \] To find the exact area, we would take the limit of this sum as \(n \to \infty\).

This example shows the explicit form of a Riemann sum using right endpoints. Calculating this limit directly can be algebraically intensive, which is why the Fundamental Theorem of Calculus is so powerful.

Definite Integrals

The limit of the Riemann sums is what we formally define as the definite integral.

Important

Definition: Definite Integral

Let \(f\) be a function on \([a,b]\). The definite integral of \(f\) from \(a\) to \(b\), denoted by \(\int_{a}^{b} f(x) \, dx\), is defined as follows: \[ \int_ {a} ^ {b} f (x) d x = \lim _ {n \to \infty} \sum_ {k = 1} ^ {n} \frac {b - a}{n} f (x _ {k} ^ {*}), \] where the limit of the Riemann sums as \(n\to \infty\) must be independent of how the sample points \(x_{k}^{*}\) are chosen.

Illustration of the definite integral as the area under a curve.

Conventions for definite integrals:

• If \(a > b\), we define \(\int_ {a} ^ {b} f (x) d x = - \int_ {b} ^ {a} f (x) d x\).
• If \(a = b\), we define \(\int_ {a} ^ {b} f (x) d x = 0\).

The definite integral gives us the exact area under the curve. The notation \(\int_{a}^{b} f(x) \, dx\) specifies the function, the variable of integration (x), and the limits of integration (a to b). The conventions for a > b and a = b are important for consistency.

Integrable Functions

Not all functions are integrable. The limit of the Riemann sum must exist.

If \(\int_{a}^{b}f(x)dx\) exists, we say that \(f\) is (Riemann) integrable on \([a,b]\).

Some Riemann Integrable Functions

Important

Theorem 6.2.3:

If \(f\) is

1. continuous,
2. monotonic, or
3. piecewise continuous with finite number of jump discontinuities

on \([a,b]\), then the definite integral \(\int_{a}^{b}f(x)dx\) exists.

Essentially, most “nice” functions we encounter in engineering and mathematics are integrable.

This theorem gives us confidence that for the types of functions we typically work with, definite integrals will exist. Continuous functions are definitely integrable. Functions that only increase or only decrease (monotonic) are also integrable. Even functions with a few breaks (piecewise continuous) can be integrable.

Definite Integral by Definition: Example

Let’s evaluate a definite integral using the Riemann sum definition.

Example 6.2.4: Find \(\int_1^3 x^2 dx\).

Solution:

1. Partition the interval: \([1, 3]\) into \(n\) subintervals of equal width, \(\Delta x = \frac{2}{n}\).

   The endpoints are \(x_{k} = 1 + k\Delta x = 1 + \frac{2k}{n}\).

   The subintervals are:

   \([ 1, 1 + \frac {2}{n} ], [ 1 + \frac {2}{n}, 1 + 2 (\frac {2}{n}) ], \dots , [ 1 + (k - 1) (\frac {2}{n}), 1 + (k) (\frac {2}{n}) ], \dots ,\)

   \(\ldots , [ 1 + (n - 1) (\frac {2}{n}), 3 ]\)

2. Choose sample points: Take \(x_k^* = x_k = 1 + \frac{2k}{n}\) (right endpoints).

3. Form the Riemann Sum:

   \(\mathrm {R i e m a n n} \mathrm {S u m} = \sum_ {k = 1} ^ {n} f (x _ {k} ^ {*}) \Delta x = \sum_ {k = 1} ^ {n} f (1 + \frac {2 k}{n}) \Delta x\)

   \(= \sum_ {k = 1} ^ {n} (1 + \frac {2 k}{n}) ^ {2} \cdot \frac {2}{n} = \frac {2}{n} \left(\sum_ {k = 1} ^ {n} (1 + \frac {4 k}{n} + \frac {4 k ^ {2}}{n ^ {2}})\right)\)

   \(= \frac {2}{n} \left(\sum_ {k = 1} ^ {n} 1 + \sum_ {k = 1} ^ {n} \frac {4 k}{n} + \sum_ {k = 1} ^ {n} \frac {4 k ^ {2}}{n ^ {2}}\right) = \frac {2}{n} \left(n + \frac {4}{n} \sum_ {k = 1} ^ {n} k + \frac {4}{n ^ {2}} \sum_ {k = 1} ^ {n} k ^ {2}\right)\)

   Using summation formulas (\(\sum k = \frac{n(n+1)}{2}\), \(\sum k^2 = \frac{n(n+1)(2n+1)}{6}\)):

   \(= \frac {2}{n} \left(n + \frac {4}{n} \cdot \frac {n (n + 1)}{2} + \frac {4}{n ^ {2}} \cdot \frac {n (n + 1) (2 n + 1)}{6}\right)\)

   \(= 2 \left(1 + 2 + \frac {2}{n} + \frac {2}{3} (2 + \frac {3}{n} + \frac {1}{n ^ {2}})\right)\)

4. Take the limit as \(n \to \infty\):

   \[ \int_ {1} ^ {3} x ^ {2} d x = \lim _ {n \to \infty} 2 \left(1 + 2 + \frac {2}{n} + \frac {2}{3} (2 + \frac {3}{n} + \frac {1}{n ^ {2}})\right) = 2 \left(1 + 2 + 0 + \frac {2}{3} (2 + 0 + 0)\right) = 2 \left(3 + \frac{4}{3}\right) = 2 \left(\frac{13}{3}\right) = \frac {2 6}{3}. \]

Remarks: Since \(x^2 \geq 0\), the value \(\int_{1}^{3} x^2 \, dx = \frac{26}{3}\) is the area of the region under the graph of \(y = x^2\), and above the \(x\)-axis, for \(1 \leq x \leq 3\).

This example is a classic demonstration of the definition. It’s quite labor-intensive! Later, the Fundamental Theorem of Calculus will allow us to calculate this much more quickly. Summation formulas are key here.

Meaning of Definite Integral

The definite integral \(\int_{a}^{b}f(x)dx\) represents the net area between the graph of \(y = f(x)\) and the \(x\)-axis.

• Parts of the graph lying above the \(x\)-axis give a positive contribution to the area.
• Parts of the graph lying under the \(x\)-axis give a negative contribution to the area.

This is because terms where \(f(x_{k}^{*}) < 0\) give a negative contribution to the Riemann sum \(\sum_{k=1}^{n} f(x_{k}^{*}) \Delta x\).

Graph showing a function both above and below the x-axis, with positive and negative areas highlighted.

It’s crucial to understand “net area.” If you need the total area, you might need to split the integral where the function crosses the x-axis and take the absolute value of the negative parts.

More on Definite Integral Meaning

Let’s clarify the definition further:

• Suppose \(f(x) \geq 0\) on \([a, b]\). The definite integral \(\int_{a}^{b} f(x) \, dx\) is the area of the region bounded below by the graph \(y = f(x)\) and above the \(x\)-axis, on \([a, b]\).

• The definite integral \(\int_{a}^{b} f(x) \, dx\) is a number which is independent of the variable \(x\). \[ \int_ {a} ^ {b} f (x) d x = \int_ {a} ^ {b} f (t) d t = \int_ {a} ^ {b} f (s) d s = \dots . \]

  The variables \(x, t, s\) are dummy variables.

Visual representation of area under a curve, where the region is entirely above the x-axis, indicating a positive definite integral.

The “dummy variable” concept is important. It means you can use any letter for the variable of integration without changing the value of the definite integral, as long as the limits of integration stay the same.

6.2.1 Properties of Definite Integrals

Definite integrals have several useful properties, similar to indefinite integrals.

Important

Theorem 6.2.5: Suppose all the definite integrals below exist. Then,

1. \(\int_{a}^{b} c \, dx = c(b - a)\). (Area of a rectangle)
2. \(\int_{a}^{b}\left(f(x)\pm g(x)\right)dx = \int_{a}^{b}f(x)dx\pm \int_{a}^{b}g(x)dx.\) (Sum/Difference Rule)
3. \(\int_{a}^{b} K f(x) \, dx = K \int_{a}^{b} f(x) \, dx\), where \(K\) is a constant. (Constant Multiple Rule)
4. \(\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx\). (Additivity of Intervals)

These properties are fundamental for manipulating and evaluating definite integrals. Property 1 is very intuitive: integrating a constant function just gives the area of a rectangle. Property 4 is useful for splitting an integral over a larger interval into smaller, more manageable parts.

Properties of Definite Integrals: Example

Let’s use these properties to evaluate an integral more easily.

Example 6.2.6: Evaluate \(\int_{1}^{3}(4 + x^{2})dx\).

Solution:

We previously evaluated \(\int_{1}^{3}x^{2}dx = \frac{26}{3}\) using the Riemann sum definition.

1. Use property 1 for the constant term:

   \(\int_{1}^{3}4dx = 4(3 - 1) = 4(2) = 8.\)

2. Use property 2 (Sum Rule):

   \(\int_ {1} ^ {3} \left(4 + x ^ {2}\right) d x = \int_ {1} ^ {3} 4 d x + \int_ {1} ^ {3} x ^ {2} d x\)

3. Combine the results:

   \(= 8 + \frac {2 6}{3}\)

   \(= \frac {24}{3} + \frac {2 6}{3} = \frac {5 0}{3}.\)

This example shows how breaking down a complex integral using these properties can simplify calculations, especially when you already know the values of simpler integrals.

Order Preserving Property

Definite integrals also respect inequalities between functions.

Important

Theorem 6.2.7: Suppose the following integrals exist and \(a < b\).

1. \(f(x)\geq 0\) on \([a,b]\Rightarrow \int_{a}^{b}f(x)dx\geq 0.\)
2. \(f(x)\geq g(x)\) on \([a,b]\Rightarrow \int_{a}^{b}f(x)dx\geq \int_{a}^{b}g(x)dx.\)
3. If \(m\leq f(x)\leq M\) on \([a,b]\), then \[ m (b - a) \leq \int_ {a} ^ {b} f (x) d x \leq M (b - a). \]

This property is quite intuitive. If a function is always positive, its net area must be positive. If one function is always above another, its integral will be greater. Property 3 is very useful for making rough estimates of integral values without actually calculating them.

Rough Estimates

Let’s use the order-preserving property for estimation.

Example 6.2.8: Estimate the value of the integral \(\int_{1}^{2}\frac{1}{x} dx\) without evaluating it.

Solution:

1. Analyze the function on the interval:

   On the interval \([1, 2]\), the function \(f(x) = 1/x\) is decreasing.

   Therefore, its largest value occurs at the left endpoint (\(x=1\)) and its smallest value occurs at the right endpoint (\(x=2\)).

   So, for \(x \in [1, 2]\):

   Minimum value \(m = f(2) = \frac{1}{2}\)

   Maximum value \(M = f(1) = 1\)

   Thus, we have \(\frac{1}{2} \leq f(x) \leq 1\) for \(x \in [1, 2]\).

2. Apply the Order-preserving property (3):

   Here \(a=1\) and \(b=2\). So \(b-a = 2-1 = 1\).

   \(m (b - a) \leq \int_ {a} ^ {b} f (x) d x \leq M (b - a)\)

   \(\frac{1}{2} (2 - 1) \leq \int_ {1} ^ {2} f (x) d x \leq 1 \cdot (2 - 1)\)

   This means:

   \(\frac {1}{2} \leq \int_ {1} ^ {2} \frac {1}{x} d x \leq 1\).

This provides a quick way to check if your calculated integral is reasonable. If you calculated an integral to be, say, 5, for this function on this interval, you’d know you made a mistake because it falls outside the estimated range.

Even Functions

Symmetry can greatly simplify definite integrals.

Note

A function \(f\) is even if \(f(-x) = f(x)\) for all \(x\).

(e.g., \(x^2, \cos x\))

Graphically, even functions are symmetric about the \(y\)-axis.

Important

Proposition 6.2.9: Suppose \(f\) is an even continuous function. Then \(\int_{-a}^{a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx\).

Example 6.2.10:

1. \(\int_{-5}^{5} x^2 \, dx = 2 \int_{0}^{5} x^2 \, dx\)
2. \(\int_{-\pi}^{\pi}\cos x dx = 2\int_{0}^{\pi}\cos x dx\)

This property is extremely useful because it can reduce the work needed to evaluate integrals over symmetric intervals. If you calculate the integral from 0 to a, you just double it for the full symmetric interval.

Odd Functions

Symmetry also plays a role for odd functions.

Note

A function \(f\) is odd if \(f(-x) = -f(x)\) for all \(x\).

(e.g., \(x^3, \sin x\))

Graphically, odd functions are symmetric about the origin.

Important

Proposition 6.2.11: Suppose \(f\) is an odd continuous function. Then \(\int_{-a}^{a} f(x) \, dx = 0\).

Example 6.2.12:

1. \(\int_{-179}^{179}x^3 dx = 0\)
2. \(\int_{-\pi}^{\pi}\sin x dx = 0\)

For odd functions over a symmetric interval, the positive area cancels out the negative area perfectly, resulting in a net area of zero. This is a great shortcut to recognize!

6.3 The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is a cornerstone of calculus. It establishes the precise inverse relationship between differentiation and the definite integral.

Newton and Leibniz exploited this relationship and used it to develop calculus into a systematic mathematical method. It is mainly used for computing areas and integrals very easily without computing them as limits of sums.

Mean Value via Definite Integral

Before diving into the FTC, let’s consider the mean value of a function.

Note

Definition 6.3.1: Mean Value of a Function

If \(f\) is continuous on \([a, b]\), then the mean value (also known as the average value) of \(f\) on \([a, b]\) is: \[ {\frac {1}{b - a}} \int_ {a} ^ {b} f (x) d x. \]

Graph illustrating the mean value of a function, showing a rectangle with the same area as that under the curve.

Example 6.3.2:

We have calculated \(\int_{1}^{3} x^{2} dx = \frac{26}{3}\).

The mean value of \(x^{2}\) on \([1,3]\) is:

\[ \frac {1}{3 - 1} \cdot \frac {2 6}{3} = \frac {1}{2} \cdot \frac {2 6}{3} = \frac {1 3}{3}. \]

The mean value of a function over an interval is essentially the height of a rectangle that would have the same area as the area under the curve over that same interval. It’s an average height.

Mean Value Theorem for Definite Integrals

This mean value concept leads to an important theorem.

Important

Theorem 6.3.3 (The Mean Value Theorem for Definite Integrals):

If \(f\) is continuous on \([a, b]\), then there is a point \(c \in [a, b]\) such that:

\[ f (c) = \frac {1}{b - a} \int_ {a} ^ {b} f (x) d x, \] i.e., \[ (b - a) f (c) = \int_ {a} ^ {b} f (x) d x. \]

Interpretation:

Suppose \(f(x) > 0\). The equation \((b - a) f (c) = \int_ {a} ^ {b} f (x) d x\) means that the area of the rectangle with height \(f(c)\) and width \((b-a)\) is equal to the area under the curve \(y=f(x)\) from \(a\) to \(b\).

In this sense, \(f(c)\) is the average value of \(f\) on the interval \([a,b]\).

This theorem guarantees that the mean value of a continuous function is actually achieved at some point c within the interval. This is similar to the Mean Value Theorem for Derivatives.

6.3.1 The First Fundamental Theorem of Calculus

This theorem establishes a direct link between differentiation and integration.

Important

Theorem 6.3.4 (The First Fundamental Theorem of Calculus):

If \(f\) is continuous on \([a, b]\), then the function \(F(x)\) defined by \[ F (x) = \int_ {a} ^ {x} f (t) d t, \quad a \leq x \leq b \] is continuous on \([a,b]\) and differentiable on \((a,b)\), and \(F^{\prime}(x) = f(x)\), i.e., \[ \frac {d}{d x} \left(\int_ {a} ^ {x} f (t) d t\right) = f (x). \]

Important Note: The lower limit of integration \(a\) is a constant, and the upper limit of integration is the variable \(x\).

This is incredibly powerful! It says that if you define a function as the accumulated area under another function, then the rate of change of that accumulated area is simply the original function itself. It truly shows integration and differentiation are inverse operations.

FTC Part 1: Examples

Let’s apply the First Fundamental Theorem of Calculus.

Example 6.3.5: Consider \(g(x) = \int_{1}^{x} \frac{\sin t}{t} \, dt\), \(1 \leq x \leq b\).

By the Fundamental Theorem of Calculus, the function \[ g (x) = \int_ {1} ^ {x} \frac {\sin t}{t} d t \] is continuous on \([1,b]\) and is differentiable on \((1,b)\).

Its derivative is given by: \[ g ^ {\prime} (x) = \frac {d}{d x} \left(\int_ {1} ^ {x} \frac {\sin t}{t} d t\right) = \frac {\sin x}{x}. \]

Example 6.3.6:

\(\frac{d}{dx}\left(\int_{2}^{x}(\sin t)\ln (t^2 +1)dt\right) = (\sin x)\ln (x^2 +1)\)

Example 6.3.7:

\(\frac{d}{dx}\left(\int_{\pi}^{x}(e^{y^2 +1})\tan^3 ydy\right) = (e^{x^2 +1})\tan^3 x\)

Example 6.3.8:

\(\frac{d}{dt}\left(\int_{179}^{t}\sqrt[3]{u^4 - 3u + 1} du\right) = \sqrt[3]{t^4 - 3t + 1}\)

These examples directly apply the theorem where the upper limit is x (or t). You simply replace the dummy variable in the integrand with the upper limit variable. The lower limit constant doesn’t affect the derivative.

FTC Part 1: Upper Limit as a Function

What if the upper limit is not just \(x\), but a function of \(x\)?

Example 6.3.9: Simplify \(\frac{d}{dx}\left(\int_x^\pi e^{(t - 3)^2}dt\right)\).

First, use the property \(\int_{a}^{b} f(x)dx = -\int_{b}^{a} f(x)dx\):

\(\int_x^\pi e^{(t - 3)^2}dt = -\int_\pi^x e^{(t - 3)^2}dt\).

Now apply FTC Part 1:

\(\frac{d}{dx}\left(-\int_\pi^x e^{(t - 3)^2}dt\right) = -e^{(x - 3)^2}\).

Example 6.3.10: \(\frac{d}{dx}\int_{1}^{\sin x}\ln (t^{2} + 1)dt\).

Solution: Note that we need to apply the chain rule in addition to the fundamental theorem of calculus.

Let \(u = \sin x\). Then the integral becomes \(\int_{1}^{u} \ln (t^2 + 1) dt\).

\(\frac {d}{d x} \int_ {1} ^ {\sin x} \ln (t ^ {2} + 1) d t = \frac {d}{d x} \int_ {1} ^ {u} \ln (t ^ {2} + 1) d t\).

Example 6.3.9 shows how to handle a variable in the lower limit. You just flip the limits and negate the integral. Example 6.3.10 sets up for the next generalization: the Chain Rule for definite integrals.

FTC Part 1 with Chain Rule

This generalizes FTC Part 1 for when the upper limit is a function of \(x\).

Important

Theorem 6.3.11: \[ \frac{d}{dx}\int_{a}^{u(x)}f(t)dt = u'(x)\cdot f\left(u(x)\right). \]

This is a powerful generalization. It’s essentially the chain rule applied to FTC Part 1. You substitute the upper limit function into the integrand, and then multiply by the derivative of that upper limit function.

FTC Part 1 with Chain Rule: Examples

Let’s put Theorem 6.3.11 into practice.

Example 6.3.12:

\(\frac{d}{dx}\int_{0}^{x^3}e^{-t^2}dt\)

Here \(u(x) = x^3\), so \(u'(x) = 3x^2\). And \(f(t) = e^{-t^2}\).

Applying the formula: \(u'(x) \cdot f(u(x)) = (3x^2) \cdot e^{-(x^3)^2}\)

\(= 3x^2 e^{-x^6}\)

Example 6.3.13: Find the first derivative of \(F(x) = \int_{x^2}^{x^3} e^{-t^2} dt\).

This involves variable limits at both top and bottom.

Use the property \(\int_{a}^{b} f(t)dt = \int_{a}^{c} f(t)dt + \int_{c}^{b} f(t)dt\).

We can write \(F(x) = \int_{x^2}^{0} e^{-t^2} dt + \int_{0}^{x^3} e^{-t^2} dt\). (Choosing \(c=0\) for convenience)

\(F(x) = -\int_{0}^{x^2} e^{-t^2} dt + \int_{0}^{x^3} e^{-t^2} dt\).

Now apply Theorem 6.3.11 to each term:

\(\frac{d}{dx}\left(-\int_{0}^{x^2} e^{-t^2} dt\right) = -(e^{-(x^2)^2} \cdot 2x) = -2x e^{-x^4}\).

\(\frac{d}{dx}\left(\int_{0}^{x^3} e^{-t^2} dt\right) = e^{-(x^3)^2} \cdot 3x^2 = 3x^2 e^{-x^6}\).

So, \(F'(x) = -2x e^{-x^4} + 3x^2 e^{-x^6}\).

Example 6.3.13 is the most general case you’ll encounter for FTC Part 1. You split the integral and apply the theorem (with the chain rule) to each part.

Remarks on Functions Defined by Integrals

It may seem odd to have functions defined via definite integral \(\int_{a}^{x}f(t)dt\). However, such functions are not uncommon in mathematics, physics, chemistry and statistics.

• The Fresnel function \(S(x) = \int_0^x \sin(\pi t^2 / 2) \, dt\) appears in Fresnel’s theory of the diffraction of light waves, and has also been applied to the design of highways.

• Another example is the sine integral function \(\mathrm{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} dt\) in electrical engineering.

• A formal way to define the natural logarithmic function is \(\ln x = \int_{1}^{x}\frac{1}{t} dt\) for \(x > 0\). From this definition, the function \(\ln x\) is one-to-one, since its derivative is \(1 / x > 0\) for \(x > 0\). Thus it has an inverse which is denoted by \(e^x\).

  From these functions, we have the formal definition of general exponential function \(a^x\) with base \(a\) defined as \(a^x = e^{x\ln a}\), its inverse function is then denoted by \(\log_a x\).

These examples highlight that defining functions as integrals isn’t just a theoretical exercise; it’s a practical tool in many scientific and engineering fields. It allows us to define functions whose antiderivatives might not be expressible in terms of elementary functions.

6.3.2 The Second Fundamental Theorem of Calculus

This is arguably the most practical part of the Fundamental Theorem. It gives us a straightforward way to evaluate definite integrals.

Important

Theorem 6.3.14 (The Second Fundamental Theorem of Calculus):

If \(f\) is continuous on \([a, b]\), then \[ \int_ {a} ^ {b} f (x) d x = G (b) - G (a), \] where \(G\) is any antiderivative of \(f\), i.e., \(G' = f\).

This theorem is why we don’t have to calculate Riemann sums anymore! It gives us a direct method to evaluate definite integrals using antiderivatives. Find any antiderivative (the constant C will cancel out), evaluate it at the upper and lower limits, and subtract.

The Evaluation Symbol

To simplify writing the evaluation of definite integrals, we use a special notation.

We define: \[ G (x) \bigg | _ {a} ^ {b} = G (b) - G (a), \] This enables us to state that if \(f\) and \(G\) are continuous on \([a, b]\) and \(G'(x) = f(x)\) on \((a, b)\), then: \[ \int_ {a} ^ {b} f (x) d x = G (x) \Bigg | _ {a} ^ {b}. \]

Using the notation \(\int f(x)dx\) for the indefinite integral, we may write: \[ \int_ {a} ^ {b} f (x) d x = \left(\int f (x) d x\right) \Bigg | _ {a} ^ {b} \]

The evaluation symbol is just a shorthand. It makes the calculation steps much clearer and more concise than writing out G(b) - G(a) every time. Remember, the constant of integration ‘C’ always cancels out in definite integrals, so you don’t need to include it when finding G(x).

FTC Part 2: Examples

Let’s use the Second Fundamental Theorem of Calculus to evaluate definite integrals.

Example 6.3.15: Evaluate \(\int_0^{\pi /2}\left(x^2 +\cos x\right)dx\). Solution: 1. Find an antiderivative \(G(x)\) of \(f(x) = x^2 + \cos x\): \(G(x) = \frac{x^3}{3} + \sin x\). (We can omit the constant \(C\)). 2. Evaluate \(G(x)\) at the limits and subtract: \[ \int_ {0} ^ {\pi / 2} \left(x ^ {2} + \cos x\right) d x = \left[ \frac{x^3}{3} + \sin x \right]_0^{\pi/2} \] \[ = \left( \frac{(\pi/2)^3}{3} + \sin(\pi/2) \right) - \left( \frac{(0)^3}{3} + \sin(0) \right) \] \[ = \left( \frac{\pi^3}{24} + 1 \right) - (0 + 0) = \frac{\pi^3}{24} + 1. \]

FTC Part 2: Examples

Example 6.3.16: Evaluate \(\int_1^2 x^{-2}dx\).

Solution:

1. Find an antiderivative \(G(x)\) of \(f(x) = x^{-2}\):

   \(G(x) = \frac{x^{-1}}{-1} = -x^{-1} = -\frac{1}{x}\).

2. Evaluate \(G(x)\) at the limits and subtract:

   \[ \int_ {1} ^ {2} x ^ {- 2} d x = \left[ -\frac{1}{x} \right]_1^2 \] \[ = \left(-\frac{1}{2}\right) - \left(-\frac{1}{1}\right) = -\frac{1}{2} + 1 = \frac{1}{2}. \]

These examples showcase the efficiency of FTC Part 2. Compare this to the laborious Riemann sum calculation for \(\int_1^3 x^2 dx\) we did earlier! This theorem simplifies things immensely.

FTC Part 2: Piecewise Functions

For piecewise functions, we use the additivity property of definite integrals.

Example 6.3.17: Evaluate the integral \(\int_{-\pi}^{\pi} f(x) \, dx\) where \[ f (x) = \left\{ \begin{array}{l l} x & \mathrm{i f} - \pi \leq x \leq 0, \\ \sin x & \mathrm{i f} 0 < x \leq \pi . \end{array} \right. \] Solution: Split the integral at \(x=0\), where the function definition changes: \[ \int_ {-\pi} ^ {\pi} f (x) d x = \int_ {-\pi} ^ {0} x d x + \int_ {0} ^ {\pi} \sin x d x. \] 1. Evaluate the first integral: \(\int_ {-\pi} ^ {0} x d x = \left[ \frac{x^2}{2} \right]_{-\pi}^0\) \(= \left(\frac{0^2}{2}\right) - \left(\frac{(-\pi)^2}{2}\right) = 0 - \frac{\pi^2}{2} = -\frac{\pi^2}{2}\). 2. Evaluate the second integral: \(\int_ {0} ^ {\pi} \sin x d x = \left[ -\cos x \right]_0^\pi\) \(= (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2\). 3. Add the results: \(\int_ {-\pi} ^ {\pi} f (x) d x = -\frac{\pi^2}{2} + 2\).

Remember that for definite integrals of piecewise functions, you integrate each piece over its respective interval and then sum the results. This relies on the additivity property of integrals.

Spot the Mistake

Let’s look at a common mistake when using FTC Part 2.

\[ \int_ {- 1} ^ {1} \frac {1}{x ^ {2}} d x = - \frac {1}{x} \Big | _ {- 1} ^ {1} = - \frac {1}{1} - \left(- \frac {1}{- 1}\right) = - 1 - 1 = - 2. \] By any meaningful definition of the integral above, the integral should be a positive number since \(1 / x^2\) is always positive.

Explanation:

We can only use the Fundamental Theorem when the integrand \(f\) is continuous on the interval of integration. The function \(1 / x^2\) is not continuous on \([-1, 1]\) because it has a vertical asymptote at \(x=0\).

Therefore, the integral above isn’t actually defined (yet) as a standard definite integral. It’s an example of an improper integral, which we will discuss later.

Caution

Always check for continuity of the integrand over the interval of integration before applying the Fundamental Theorem of Calculus. Failure to do so can lead to incorrect results!

6.3.3 An Application of Fundamental Theorem of Calculus

The FTC allows us to evaluate limits of Riemann sums more easily.

Example 6.3.18: Evaluate \(\lim_{n\to \infty}\frac{\pi}{n}\left(\sin \frac{\pi}{n} +\sin \frac{2\pi}{n} +\ldots +\sin \frac{n\pi}{n}\right)\).

Solution: We shall express the limit as a definite integral. Recall the definition: \[ \int_ {a} ^ {b} g (x) d x = \lim _ {n \rightarrow \infty} \sum_ {k = 1} ^ {n} \Delta x \cdot g \left(x _ {k} ^ {*} \right) \] where \(\Delta x = \frac{b-a}{n}\).

Rewrite the given sum: \[ {\frac {\pi}{n}} \left(\sin {\frac {\pi}{n}} + \sin {\frac {2 \pi}{n}} + \ldots + \sin {\frac {n \pi}{n}}\right) \] \[ = \sum_ {k = 1} ^ {n} \frac {\pi}{n} \sin \left(\frac {k \pi}{n}\right) = \sum_ {k = 1} ^ {n} \frac {1}{n} \underbrace{\left( \pi \sin \left(\left(\frac {k}{n}\right) \pi\right)\right)}_{g (x _ {k} ^ {*})} \] Comparing this to the Riemann sum form, we can identify: - \(\Delta x = \frac{1}{n}\), so \(b-a = 1\). Let’s assume the interval is \([0,1]\), so \(a=0, b=1\). - \(x_{k}^{*} = \frac{k}{n}\) (right endpoint choice for \(x_k = a + k\Delta x = 0 + k\frac{1}{n}\)). - \(g(x) = \pi \sin (\pi x)\).

6.3.3 An Application of Fundamental Theorem of Calculus

Thus, the limit is equal to the definite integral: \[ \lim _ {n \rightarrow \infty} \sum_ {k = 1} ^ {n} \frac {\pi}{n} \sin \frac {k \pi}{n} = \int_ {0} ^ {1} \pi \sin (\pi x) d x. \] Now, evaluate the integral using FTC Part 2: \[ = \pi \left[ -\frac {1}{\pi} \cos (\pi x) \right]_0^1 = - \cos (\pi x) \Big | _ {0} ^ {1} \] \[ = (-\cos (\pi \cdot 1)) - (-\cos (\pi \cdot 0)) = (-\cos \pi) - (-\cos 0) \] \[ = (-(-1)) - (-1) = 1 + 1 = 2. \]

This is a very powerful application. Recognizing a limit as a Riemann sum allows you to use the FTC to evaluate it directly, bypassing complex summation formulas. The key is to correctly identify \(\Delta x\), \(f(x_k^*)\), and the interval \([a,b]\).