Notes for Calculus

Chapter 4: Integration

Author: S.K. Cheng

Table of Contents

• Notes for Calculus
  • Chapter 4: Integration
    • Author: S.K. Cheng
  • Table of Contents
    • 4.1 Indefinite Integrals
      • 4.1.1 Antiderivative, Integrand, and Variable of Integration
      • 4.1.2 Basic Integration Formulas
      • 4.1.3 Properties of Indefinite Integrals
      • Problems for this section
    • 4.2 Integration by Substitution and by Parts
      • Problems for this section
    • 4.3 Definite Integrals
      • Exercises for this section
    • 4.4 Properties of Definite Integrals
      • Problems for this section
    • 4.5 The Connection between Definite and Indefinite Integrals
      • Problems for this section
    • 4.6 Area of a Plane Region
      • Problems for this section
    • 4.7 First-Order Differential Equations
      • Problems for this section
    • 4.8 Using Python to Solve Integration and Differential Equations
      • Problems for this section
    • References

4.1 Indefinite Integrals

4.1.1 Antiderivative, Integrand, and Variable of Integration

Recall what we have learned in the previous chapter, we know that the derivative of a function is a function that gives the rate of change of the original function. In this chapter, we will learn the opposite of differentiation.

Given a function $f(x)$ defined on an interval $I$, suppose there is a function $F(x)$ defined in $I$ such that

$$F'(x) = \frac{dF(x)}{dx} = f(x)$$

for all $x$ in $I$. Then, we say that $F(x)$ is an antiderivative of $f(x)$ on $I$. Now, consider finding the antiderivative of a function $f(x) = x^2$.

By what we have learned in differentiation, we know that the derivative of $x^3$ is $3x^2$. Therefore, we can say that $\frac{1}{3}x^3$ is an antiderivative of $x^2$. However, one may propose another question: is $\frac{1}{3}x^3 + 1$ also an antiderivative of $x^2$? The answer is yes, because the derivative of a constant is zero. In other words, the antiderivative of a function is not unique. In general, if $F(x)$ is an antiderivative of $f(x)$, then $F(x) + C$ is also an antiderivative of $f(x)$, where $C$ is a constant.

Now, we introduce the concept of indefinite integral. The definition is as follows:

Definition 4.1.1 If $F(x)$ is an antiderivative of $f(x)$ on an interval $I$, then the expression $F(x)+C$ involving arbitrary constant $C$ is called an indefinite integral of $f(x)$ on $I$, and is denoted by

$$\int f(x)dx = F(x) + C.$$

Remark: The symbol $\int$ is called the integral sign, the operation leading from the function $f(x)$, called integrand, to the indefinite integral $F(x) + C$ is called integration, and the variable $x$ is called the variable of integration.

Example 4.1.1 Clearly, by differentiation, we have

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

In fact, if $x \in (0, +\infty)$, then

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

Theorem 4.1.1 The following formulas hold:

$$\begin{aligned} \frac{d}{dx}\int f(x)dx &= f(x), \\ d\int f(x)dx &= f(x)dx, \\ \int f'(x)dx &= f(x) + C. \end{aligned}$$

Proof: If $F(x)$ is any antiderivative of $f(x)$ (in some underlying interval), then

$$\begin{aligned} \frac{d}{dx}\int f(x)dx &= \frac{d}{dx}(F(x) + C) = F'(x) = f(x),\\ d\int f(x)dx &= d(F(x) + C) = F'(x)dx = f(x)dx,\\ \int f'(x)dx &= F(x) + C. \end{aligned}$$

Remark: The fact that $df(x) = f'(x) dx$ suggests the definition

$$\int df(x) = \int f'(x)dx.$$

It then follows from Theorem 4.1.1 that

$$\int df(x) = \int f'(x)dx = f(x) + C.$$

By comparison, one may observe that the symbols $d$ and $\int$ cancel each other when they are used together. And therefore, we may deduce an integration formula from any given differentiation formula.

For example, we have $d(-\cos x) = \sin x dx$, and hence

$$\int \sin x dx = \int (-\cos x)' dx = -\cos x + C.$$

The formula

$$(\frac{1}{a}\arctan \frac{x}{a})' = \frac{1}{a} \frac{a}{a^2 + x^2} = \frac{1}{a^2 + x^2}$$

implies that

$$\int \frac{1}{a^2 + x^2}dx = \frac{1}{a}\arctan \frac{x}{a} + C.$$

and so on.

By continuing this process, we can obtain a large number of integration formulas from the differentiation formulas we have learned. In the following, we will list some of the most important integration formulas.

4.1.2 Basic Integration Formulas

1. $\int x^n dx = \frac{1}{n+1}x^{n+1} + C$, where $n \neq -1$.
2. $\int e^x dx = e^x + C$.
3. $\int a^x dx = \frac{1}{\ln a}a^x + C$.
4. $\int \sin x dx = -\cos x + C$.
5. $\int \cos x dx = \sin x + C$.
6. $\int \frac{1}{x}dx = \ln|x| + C$.
7. $\int \frac{1}{a^2 + x^2}dx = \frac{1}{a}\arctan \frac{x}{a} + C$.
8. $\int \frac{1}{\sqrt{a^2 - x^2}}dx = \arcsin \frac{x}{a} + C$.
9. $\int \frac{1}{\sqrt{x^2 + a^2}}dx = \ln(x + \sqrt{x^2 + a^2}) + C$.
10. $\int \frac{1}{\sqrt{x^2 - a^2}}dx = \ln|x + \sqrt{x^2 - a^2}| + C$.
11. $\int \frac{dx}{\sin^2 x} = \int \csc^2 x dx = -\cot x + C$.
12. $\int \frac{dx}{\cos^2 x} = \int \sec^2 x dx = \tan x + C$.

4.1.3 Properties of Indefinite Integrals

We know that if two functions' derivatives are the same, then the two functions differ by a constant. Then one may wonder if the derivatives of two functions are the same, then are the integrals of the two functions the same?

To view this, we note that

$$\begin{aligned} \frac{d}{dx}\int f(x)dx &= \frac{d}{dx}\int g(x)dx,\\ \frac{d}{dx}(\int f(x)dx - \int g(x)dx) &= 0,\\ \int f(x)dx - \int g(x)dx &= C,\\ \int f(x)dx &= \int g(x)dx + C. \end{aligned}$$

But indefinite integrals are defined only to within an arbitrary constant $C$, and in this sense, the integrals of two functions differing by a constant are the same.

The following properties of indefinite integrals are immediate consequences of the definition of the indefinite integral:

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

Corollary 4.1.1 If $f_1(x),f_2(x),\cdots,f_n(x)$ all have antiderivatives on an interval $I$, and if $c_1,c_2,\cdots,c_n$ are constants, then

$$\int (c_1f_1(x) + c_2f_2(x) + \cdots + c_nf_n(x))dx = c_1\int f_1(x)dx + c_2\int f_2(x)dx + \cdots + c_n\int f_n(x)dx.$$

The proof is trivial as one may apply the properties of indefinite integrals and obtain the result.

Another useful result is

Theorem 4.1.2 If

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

then

$$\int f(ax + b)dx = \frac{1}{a}F(ax + b) + C.$$

for arbitrary constants $a$ and $b$ where $a \neq 0$. In particular,

$$\int f(x+b)dx = F(x+b) + C,\\ \int f(ax)dx = \frac{1}{a}F(ax) + C.$$

Proof: By hypothesis, we have

$$F'(x) = f(x).$$

Then, by the chain rule, we have

$$(F'(ax+b))' = aF'(ax+b) = af(ax+b).$$

Therefore, we have

$$\int f(ax+b)dx = F(ax+b) + C.$$

We now use this new technique to evaluate a few integrals.

Example 4.1.2 Evaluate

$$\int (6x^2-3x+2)dx.$$

Solution: We have

$$\begin{aligned} \int (6x^2-3x+2)dx &= \int 6x^2dx - \int 3xdx + \int 2dx\\ &= 6\int x^2dx - 3\int xdx + 2\int dx\\ &= 6\frac{1}{3}x^3 - 3\frac{1}{2}x^2 + 2x + C\\ &= 2x^3 - \frac{3}{2}x^2 + 2x + C. \end{aligned}$$

Example 4.1.3: Evaluate

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

Solution: By observation, this integral can be rewritten as

$$\begin{aligned} \int \frac{dx}{\sin^2 x \cos^2 x} &= \int \frac{(\sin^2 x + \cos^2 x)dx}{\sin^2x \cos^2x}\\ &= \int \frac{dx}{\sin^2 x} + \int \frac{dx}{\cos^2 x}\\ &= -\cot x - \tan x + C. \end{aligned}$$

Example 4.1.4 Evaluate

$$\int \cos^2 xdx.$$

Solution: Since $\cos^2 x = \frac{1}{2}(1 + \cos 2x)$, we have

$$\begin{aligned} \int \cos^2 xdx &= \frac{1}{2}\int (1 + \cos 2x)dx\\ &= \frac{1}{2}\int dx +\frac{1}{2} \int \cos 2xdx\\ &= \frac{1}{2}x + \frac{1}{4}\sin 2x + C. \end{aligned}$$

Problems for this section

1. Evaluate
   1. $\int (x^4-3x^2+2)dx$.
   2. $\int x^2(5-x)^4dx$.
   3. $\int (1-x)(1-2x)(1-3x)dx$.
   4. $\int (\frac{1-x}{x})^2 dx$.
2. Prove that

$$\int \frac{dx}{1+\sin x} = -\tan(\frac{\pi}{4}-\frac{x}{2}) + C$$

3. Prove that

$$\int \frac{dx}{(x+a)(x+b)} = \frac{1}{a-b}\ln|\frac{x+a}{x+b}| + C \quad (a \neq b).$$

4.2 Integration by Substitution and by Parts

Given that

$$\int g(t)dt = G(t) + C \quad (G'(t) = g(t))$$

let $t = t(x)$ be a differentiable function of $x$ in an interval $I$ such that $g(t(x))$ is defined and continuous on $I$. Then, by the chain rule, we have

$$\frac{d}{dx}G(t(x)) = G'(t(x))t'(x) = g(t(x))t'(x),$$

which implies that

$$\int g(t(x))t'(x)dx = G(t(x)) + C.$$

This is the substitution rule for integration.

Suppose the integral

$$\int f(x)dx$$

can be recognized as being of the form

$$\int g(t(x))t'(x)dx$$

where $t = t(x)$ is a differentiable function of $x$. Moreover, we can integrate $g(t)$, obtaining

$$\int g(t)dt = G(t) + C.$$

Then, we can evaluate the integral of $f(x)$

$$\int f(x)dx = \int g(t(x))t'(x)dx = G(t(x)) + C.$$

Now we try to use the substitution rule to evaluate the integral

Example 4.2.1 Evaluate

$$\int \sin^3 x \cos x dx.$$

Solution: Since we can have $d(sin x) = \cos x dx$, we can let $t = \sin x$, then $dt = \cos x dx$. Therefore, we have

$$\begin{aligned} \int \sin^3 x \cos x dx &= \int t^3 dt\\ &= \frac{1}{4}t^4 + C\\ &= \frac{1}{4}\sin^4 x + C. \end{aligned}$$

Example 4.2.2 Evaluate

$$\int \sqrt{a^2-x^2}dx.$$

Solution: First, we observe that this integral contains a square root of a quadratic polynomial. One may think of using trigonometric functions to simplify since we have the identity

$$\sin^2 x + \cos^2 x = 1.$$

Therefore, we let $x = a\sin t$, then $dx = a\cos tdt$. Therefore, we have

$$\begin{aligned} \int \sqrt{a^2-x^2}dx &= \int \sqrt{a^2-a^2\sin^2 t}a\cos tdt\\ &= \int a^2\cos^2 tdt\\ &= \int \frac{a^2(1+\cos 2t)}{2}dt\\ &= \frac{a^2}{2}t + \frac{a^2}{4}\sin 2t + C\\ &= \frac{a^2}{2}\arcsin \frac{x}{a} + \frac{1}{2}a\sin t a \cos t + C\\ &= \frac{a^2}{2}\arcsin \frac{x}{a} + \frac{1}{2}x\sqrt{a^2-x^2} + C. \end{aligned}$$

We now turn to another important technique of integration. Let $u = u(x)$ and $v = v(x)$ be differentiable functions of $x$ such that $u'(x)v(x)$ and $u(x)v'(x)$ have antiderivatives. Then, by the product rule, we have

$$d(uv) = udv + vdu$$

and hence

$$udv = d(uv) - vdu$$

which implies that

$$\int udv = uv - \int vdu.$$

This is the integration by parts formula. However, one may wonder how does it relate to our integration problems. Let us consider the following example.

Example 4.2.3 Evaluate

$$\int x\sin xdx.$$

Solution: We can let $u = x$ and $dv = \sin xdx$, then $du = dx$ and $v = -\cos x$. Therefore, we have

$$\begin{aligned} \int x\sin xdx &= -x\cos x - \int -\cos xdx\\ &= -x\cos x + \sin x + C. \end{aligned}$$

Remark: One may wonder why we choose $u = x$ and $dv = \sin xdx$. The reason is that unlike $\sin x$, $x$ is simplified by differentiation. Therefore, we choose $u = x$ and $dv = \sin xdx$. The whole point is to make $\int vdu$ easier to integrate than $\int udv$.

Example 4.2.4 Evaluate

$$\int \ln xdx.$$

Solution: We can let $u = \ln x$ and $dv = dx$, then $du = \frac{1}{x}dx$ and $v = x$. Therefore, we have

$$\begin{aligned} \int \ln xdx &= x\ln x - \int x\frac{1}{x}dx\\ &= x\ln x - \int dx\\ &= x\ln x - x + C. \end{aligned}$$

Remark: This example is to show that even we only have a logarithmic function, we can still use integration by parts to evaluate the integral.

Problems for this section

1. Use integration by substitution to evaluate
   1. $\int e^{x^2}xdx$.
   2. $\int \frac{xdx}{1+x^4}$.
   3. $\int \frac{\ln x dx}{x}$.
2. Use integration by parts to evaluate
   1. $\int x\cos xdx$.
   2. $\int x^3 \ln x dx$.
   3. $\int \sqrt{x} \ln x dx$.

4.3 Definite Integrals

The definite integral is one of the key concepts of Calculus, and is habitually encountered in various applications. Rigoriously, the definite integral we discussed in this section is the Riemann integral. We first begin by studying the following example.

Example 4.3.1: Given a particle moving along a straight line, let $v=v(t)$ be the particle's velocity at time $t$. Find the distance traveled by the particle from time $a$ to time $b$.

Solution: Assuming that the velocity is continuous (One may note that the physics meaning of velocity guarantees that the velocity is continuous), we divide the interval $[a,b]$ into $n$ small subintervals by introducing a large number of points of subdivision $t_1,t_2,\cdots,t_{n-1}$ such that $a = t_0 < t_1 < t_2 < \cdots < t_{n-1} < t_n = b$, where, in the interest of a uniform notation, the end points $a$ and $b$ of the original interval are assigned alternative symbols $t_0$ and $t_n$, as if they were points of subdivision too. Let

$$\Delta t_i = t_i - t_{i-1} \quad (i = 1,2,\cdots,n),$$

and let $\lambda$ be the maximum length of the subintervals, i.e., let

$$\max\{t_1-t_0,t_2-t_1,\cdots,t_n-t_{n-1}\} = \max\{\Delta t_1,\Delta t_2,\cdots,\Delta t_n\} = \lambda.$$

Being continuous, the velocity does not change abruptly, and hence, it is a reasonable assumption that the velocity is approximately constant over each subinterval which we denoted by $v(\tau_i)$ during $[t_{i-1},t_i]$, where $\tau_i$ is a point in $[t_{i-1},t_i]$. Then, the distance traveled by the particle during $[t_{i-1},t_i]$ is approximately

$$v(\tau_i)(t_i - t_{i-1}) = v(\tau_i)\Delta t_i,$$

and hence the distance traveled during the whole interval $[a,b]$ is approximately

$$\sum_{i=1}^n v(\tau_i)\Delta t_i.$$

This approximation gets better as the subintervals $[t_0,t_1],[t_1,t_2],\cdots,[t_{n-1},t_n]$ get jointly smaller. Therefore, the distance traveled by the particle from time $a$ to time $b$ is defined to be

$$\lim_{\lambda \to 0} \sum_{i=1}^n v(\tau_i)\Delta t_i.$$

Remark: Let $x=x(t)$ be the particle's poisition at time $t$. Then, since

$$v(t) = \frac{dx(t)}{dt},$$

by the very definition of velocity, $x(t)$ must be an antiderivative of $v(t)$. On the other hand, in terms of $x(t)$, the distance $I$ (which we denoted) traveled by the particle from time $a$ to time $b$ is obviously $x(b) - x(a)$. Therefore, we have

$$I = \lim_{\lambda \to 0} \sum_{i=1}^n v(\tau_i)\Delta t_i = x(b) - x(a).$$

This is indeed true and is an immediate consequence of the very important theorem we are about to discuss, which is the Fundamental Theorem of Calculus.

Example 4.3.2 Find the area $A$ of the plane region bounded by the lines $x=a$ and $x=b$, the $x-$axis and the curve

$$y=f(x) \quad (a \leq x \leq b),$$

where $f(x) \geq 0$

Solution: To illustrate, one may think of the following figure:

Then we may think of the regions as a kind of trapezoid with three straight sides and on curved side. Such regions are not considered in elementary geometry. Hence we need to find a way to calculate the area of such regions. As in the preceding example, we divide the interval $[a,b]$ into $n$ subintervals by introducing points of subdivision $x_0,x_1,\cdots,x_n$ such that $a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b$. Let

$$\Delta x_i = x_i - x_{i-1} \quad (i = 1,2,\cdots,n),$$

and let $\lambda$ be the maximum length of the subintervals, i.e., let

$$\lambda = \max\{\Delta x_1,\Delta x_2,\cdots,\Delta x_n\}.$$

Then we may have the following figure:

In the figure, we divide the region into $n$ subintervals and approximate the area of each subinterval by a rectangle. Then, the total area of the region is approximately

$$\sum_{i=1}^n f(\xi_i)\Delta x_i,$$

where $\xi_i$ is a point in $[x_{i-1},x_i]$. This approximation gets better as the subintervals get smaller. Therefore, the area of the region we denoted by $A$ is defined to be

$$A = \lim_{\lambda \to 0} \sum_{i=1}^n f(\xi_i)\Delta x_i.$$

The existence of the limit is guaranteed by the continuity of $f(x)$.

Moreover, many other geometrical and pphysical problems lead to expressions of the same kind. They are all instances of the key concept of a definite integral:

Definition 4.3.1 Given a function $f(x)$ defined in an interval $[a,b]$, let $x_1,\dots,x_{n-1}$ be points of subdivision such that

$$a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b,$$

and let $\xi_i$ be an arbitrary point of the subinterval $[x_{i-1},x_i]$, of length $\Delta x_i = x_i - x_{i-1}$. Suppose the sum

$$\sigma = \sum_{i=1}^n f(\xi_i)\Delta x_i$$

approaches a (finite) limit as

$$\lambda = \max\{\Delta x_1,\Delta x_2,\cdots,\Delta x_n\} \to 0.$$

Then, this limit is called the definite integral of $f(x)$ from $a$ to $b$, and is denoted by

$$\int_a^b f(x)dx.$$

and the function $f(x)$ is said to be integrable on $[a,b]$.

Remark: Since definite integration is an operation producing a number from a given function $f(x)$, the symbol chosen for the variable of integration is unimportant, and any other symbol would do as well. Thus

$$\int_a^b f(x)dx = \int_a^b f(t)dt = \int_a^b f(u)du = \cdots.$$

Theorem 4.3.1 If $f(x)$ is continuous on $[a,b]$, then $f(x)$ is integrable on $[a,b]$.

Exercises for this section

1. Write an expression for the area of the figure bounded by the curve $y=\ln x$ and the lines $x=3,x=5,y=1$.
2. Suppose a pot of water is heated up at the rate of $f(t)$ degrees per minute. How much does the temperature of the water change after $10$ minutes if $f(t) = 2t^2 - 3t + 1$?

4.4 Properties of Definite Integrals

Since we have discussed the definition of definite integrals, we now turn to the properties of definite integrals.

Theorem 4.4.1 If $f$ is continuous in $[a,b]$, then

$$k\int_a^b f(x)dx = \int_a^b kf(x)dx$$

where $k$ is a constant.

You may prove by using definition of definite integrals.

Theorem 4.4.2 If $f$ and $g$ are continuous in $[a,b]$, then

$$\int_a^b (f(x) + g(x))dx = \int_a^b f(x)dx + \int_a^b g(x)dx.$$

Corollary 4.4.1 If $f$ and $g$ are continuous in $[a,b]$, and $k$ is arbitrary constant, then

$$\int_a^b (kf(x) + g(x))dx = k\int_a^b f(x)dx + \int_a^b g(x)dx.$$

Theorem 4.4.3 If $f$ is continuous in $[a,b]$, and if $c$ is an interior point of $[a,b]$, then

$$\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx.$$

This Theorem has a more general case:

Theorem 4.4.4 If $f$ is continuous an interval containing the points $a,b,c$, then

$$\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx.$$

Remark: The Theorem 4.4.4 is a generalization of Theorem 4.4.3. In this theorem, the point $c$ is not necessarily an interior point of $[a,b]$.

Theorem 4.4.5 (Mean Value Theorem for Integrals) If $f$ is continuous in $[a,b]$, then there exists a point $c$ in $[a,b]$ such that

$$\int_a^b f(x)dx = f(c)(b-a).$$

Remark: By moving the terms, we have the number

$$\frac{1}{b-a}\int_a^b f(x)dx$$

which is the average value of $f(x)$ in $[a,b]$.

Problems for this section

1. Prove that if $f$ is continuous in $[a,b]$, and if $m \leq f(x) \leq M$ for all $x$ in $[a,b]$, then

$$m(b-a) \leq \int_a^b f(x)dx \leq M(b-a).$$

2. Let $f$ be continuous and nonnegative in $[a,b]$, and suppose

$$\int_a^b f(x)dx = 0.$$

Prove that $f(x) = 0$ for all $x$ in $[a,b]$. 3. Determine that which integral is larger:

1. $\int_0^1 xdx$ or $\int_0^1 x^2dx$.
2. $\int_0^{\pi/2}xdx$ or $\int_0^{\pi/2}\sin xdx$.

4.5 The Connection between Definite and Indefinite Integrals

According to Theorem, if $f(x)$ is continuous in $[a,b]$, then $f(x)$ has a definite integral in $[a,b]$. We now prove a similar result for indefinite integrals:

Theorem 4.5.1 If $f(x)$ is continuous in $[a,b]$, then the function

$$\Phi(x) = \int_a^x f(t)dt$$

is an antiderivative of $f(x)$ in $[a,b]$. In particular, $f(x)$ has an indefinite integral $\Phi(x) + C$ in $[a,b]$.

Proof: The existence follows from Theorem 4.3.1. Suppose $x$ and $x+\Delta x$ both belong to $[a,b]$. Then,

$$\begin{aligned} \Phi(x+\Delta x) - \Phi(x) &= \int_a^{x+\Delta x}f(t)dt - \int_a^x f(t)dt\\ &= \int_x^{x+\Delta x}f(t)dt. \end{aligned}$$

By the Mean Value Theorem for Integrals, there exists a point $c$ in $[x,x+\Delta x]$ such that

$$\int_x^{x+\Delta x}f(t)dt = f(c)\Delta x.$$

Therefore, we have

$$\Phi(x+\Delta x) - \Phi(x) = f(c)\Delta x.$$

By taking the limit as $\Delta x \to 0$, then $f(\xi) \to f(x)$, we have

$$\Phi'(x) = f(x).$$

Therefore, $\Phi(x)$ is an antiderivative of $f(x)$ in $[a,b]$.

Remark: The content of this theorem can be written concisely as

$$\frac{d}{dx}\int_a^x f(t)dt = f(x),$$

or

$$\int f(x)dx = \int_a^x f(t)dt + C.$$

As its name implies, the next theorem is particularly important:

Theorem 4.5.2 (Fundamental Theorem of Calculus) If $f(x)$ is continuous in $[a,b]$, then

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

where $F(x)$ is any antiderivative of $f(x)$ in $[a,b]$.

Proof: By Theorem 4.5.1, the function

$$\Phi(x) = \int_a^x f(t)dt$$

is an antiderivative of $f(x)$ in $[a,b]$. Therefore, it follows that

$$\Phi(x) = F(x) + C$$

where $F(x)$ is any other antiderivative of $f(x)$ in $[a,b]$. However, one may note that

$$\Phi(a) = F(a) + C = 0,$$

also since obviously

$$\Phi(b) = \int_a^b f(t)dt = \int_a^b f(x)dx,$$

and

$$\Phi(b) = F(b) + C = F(b) - F(a),$$

we have

$$\int_a^b f(x)dx = F(b) - F(a).$$

Remark: Note that the Fundamental Theorem of Calculus is a powerful tool for evaluating definite integrals. It allows us to evaluate definite integrals by finding antiderivatives of the integrand.

Example 4.5.1 Given a particle moving along a straight line, let $x(t)$ and $v(t)$ be the particle's position and velocity at time $t$. Then the distance $l$ traversed by the particle between the times $t=a$ and $t=b$ is given by

$$l = \int_a^b v(t)dt = \int_a^b \frac{dx(t)}{dt}dt = \int_a^b x'(t)dt = x(b) - x(a).$$

Example 4.5.2 Evaluate

$$\int^b_a x^rdx$$

where $r \neq -1$.

Solution: We have

$$\int^b_a x^rdx = \int^b_a x^rdx= \frac{1}{r+1}x^{r+1}\Big|_a^b = \frac{1}{r+1}(b^{r+1} - a^{r+1}).$$

Remark: For evaluating definite integrals, one may use the Fundamental Theorem of Calculus to find antiderivatives of the integrand, and then substitute the limits of integration into the antiderivative. Also the symbol $\Big|_a^b$ is used for the evaluation of the antiderivative at the limits of integration.

Problems for this section

1. Evaluate
   1. $\int^3_1 x^3dx$.
   2. $\int^{\pi/2}_0 \cos xdx$.
   3. $\int^3_0 e^{\frac{x}{3}}dx$.
2. Recall what we have learned in Integration by Substitution and by Parts, and evaluate
   1. $\int^8_3 \frac{x}{\sqrt{1+x}}dx$;
   2. $\int^a_0 x^2\sqrt{a^2-x^2}dx \quad (a>0)$.
3. Prove that if $f$ is continuous in $[0,1]$, then

   $$\int^{\frac{\pi}{2}}_0 f(\sin x)dx = \int^{\frac{\pi}{2}}_0 f(\cos x)dx.$$

   Show that

   $$\int^{\frac{\pi}{2}}_0 \cos^n xdx = \int^{\frac{\pi}{2}}_0 \sin^n xdx \quad (n \in \mathbb{N}).$$

4.6 Area of a Plane Region

It will be recalled that the integral $\int^b_a f(x)dx$ is the only reasonable way of defining the area "under the curve $y=f(x)$ from $a$ to $b$" more precisely, the area of the region bounded by the curve $y=f(x)$, the $x-$axis, and the lines $x=a$ and $x=b$. We now turn to the problem of finding the area of a plane region bounded by two curves.

The area of the plane region bounded by the lines $x = a$ and $x = b$, the $x-$axis, and the curves $y = f(x)$ and $y = g(x)$, where $f(x) \geq g(x)$ in $[a,b]$, is defined to be

$$\int^b_a (f(x) - g(x))dx.$$

Remark: The area of the region bounded by the curves $y=f(x)$ and $y=g(x)$ in $[a,b]$ is the difference between the areas under the two curves.

Example 4.6.1 Find the area of the region bounded by the curves $y=\sqrt{x}$ and $y=x^2$.

Solution: The region is shown in the following figure:

The only problem is how to find the points of intersection of the two curves when we are given the equations of the curves. Since one may note that the intersection of the two curves is only in the first quadrant, we have

$$\begin{aligned} \sqrt{x} &= x^2,\\ x &= 1. \end{aligned}$$

We now have the intersection point $(1,1)$. But how can we determine the area of the region? We may see from the Figure 3 that the area of the region is the difference between the areas under the two curves. Therefore, we have

$$\begin{aligned} Area &= \int^1_0 (\sqrt{x} - x^2)dx\\ &= \int^1_0 (x^{\frac{1}{2}} - x^2)dx\\ &= \frac{2}{3}x^{\frac{3}{2}} - \frac{1}{3}x^3\Big|_0^1\\ &= \frac{2}{3} - \frac{1}{3}\\ &= \frac{1}{3}. \end{aligned}$$

Remark: For student, one may wonder how to know exactly which curve is above the other. The answer is that one should plot the two curves and see which curve is above the other in the region of interest since the area bounded should be positive.

Problems for this section

1. What is the area between the line $x+y=2$ and the curve $y=x^2$?
2. What is the area bounded by the $x-$axis and the curve $y=\sin x$ from $x=0$ to $x=\pi$?
3. Find the area of the region bounded by the curves $y=\ln x$ and $y=\ln^2 x$.

4.7 First-Order Differential Equations

By a differential equation, we mean an equation involving at least one derivative

$$y' = f'(x), \quad y'' = f''(x), \quad \cdots$$

of a function $y = f(x)$ and often the function itself. A differential equation is said to be of order $n$ if $n$ is the highest order of the derivatives appearing in the equation. Thus

$$2xy' - 5y = 0, \quad y'' + y = 0$$

are first-order and second-order differential equations, respectively. This section will be devoted entirely to first-order differential equations. The general form of such an equation is

$$F(x,y,y')=0,$$

where $F$ is a function of three variables $x,y,y'$. By a solution of the differential equation, we mean any function $y=\phi(x)$ such that

$$F(x,\phi(x),\phi'(x)) = 0,$$

holds identically. For example, the function $y = x^2$ is a solution of the differential equation

$$xy' - 2y = 0.$$

since

$$x(x^2)' - 2x^2 = x(2x) - 2x^2 = 0.$$

The solution of a differential equation is not unique. For example, the function $y = 0$ is also a solution of the differential equation above. Therefore, the solution

$$y = Cx^2$$

is a general solution of the differential equation, where $C$ is an arbitrary constant.

Example 4.7.1 The simplest first_order differential equation is

$$y' = f(x),$$

and its general solution is just the indefinite integral of $f(x)$:

$$y = \int f(x)dx + C.$$

Example 4.7.2 Solve the differential equation