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3 Integrals

The integral of a function f(x) is usually thought of as the area between the graph of f(x) and the x-axis. Although this is not a false idea, this description is only a piece of a much larger puzzle. It fails to unlock the true potential of integrals and prohibits a deeper understanding of what integrals are.

Integrals are, in essence, multiplication [5].
When we multiply, we are using repeated addition with numbers that remain fixed. For example, 4\cdot 30 is adding 30 fours together (4+4+...+4). But what happens if the four is changing? What if we had this:

    \[4+4^2+4^3+...+4^{30}?\]

Here is where integration comes in.

Azad uses the example of \text{distance}=\text{speed}\cdot \text{time} [5]. In real life, the speed will mostly likely be changing and will be different at different points in time. So, the equation would look like this:

    \[\text{distance}=\int s(t)\,\,dt ,\]

where s(t) is the speed at time t. First, the integral breaks the total time into many tiny intervals. Then, the speed at time t, where t is the beginning of the interval, is multiplied by the length of the interval, dt, resulting in a tiny distance. This is done for each interval which results in a collection of tiny distances. Finally, the distances are added together to get the total distance [5]. All of this is done with this single equation:

    \[\text{distance}=\int s(t)\,\,dt.\]

Now let’s look at the formal definition provided by Larson and Edwards [15].




Definition I.7

A function F is an antiderivative of f on an interval I when F'(x)=f(x) for all x in I.


In other words, 


      \[\int f(x) \, dx=F(x).\]


It is important to understand that the definition says F is an antiderivative of f [15]. A function f can have many antiderivatives. The reasoning behind this goes back to the constant rule where 


      \[\frac{d}{dx}[c]=0\]


 for some real number c. Consider f(x)= 2x. Some antiderivatives include 


      \[F(x)=x^2, F(x)=x^2+5,\,\,\text{and}\,\, F(x)=x^2-5,764.\]


 When we take the derivative of any of these, we end up with f(x)= 2x because the constant always becomes 0. For this reason, we say the antiderivative of f(x) is 


      \[F(x)=x^2+ C\]


 where C is a real number constant.


Just as we had derivative rules, we also have integral rules. A few of the basic ones are listed by Larson and Edwards as follows [15].


    

  1. \displaystyle\int 0 \,dx=C
    2. 

  2. \displaystyle\int k\, dx=kx+C
    3. 

  3. \displaystyle\int kf(x) dx = k\displaystyle\int f(x) dx, where k is a constant
    4. 

  4. \displaystyle\int x^n dx = \displaystyle\frac{x^{n+1}}{n+1}+C, n \neq -1
    5. 

  5. \displaystyle\int [f(x)\pm g(x)]\, dx = \displaystyle\int f(x)\, dx \pm \displaystyle\int g(x)\, dx
    6. 



Let’s now work through an example to demonstrate some of these rules. This problem is one I completed for Calculus I and comes from Larson and Edwards [15].




Example 8

Find the antiderivative of f(x)=\displaystyle\frac{x^4-3x^2+5}{x^4}.


Solution


      \begin{align*} \int f(x) \,dx&=\int \frac{x^4-3x^2+5}{x^4} \,dx\\ &=\int 1-3x^{-2}+5x^{-4}\, dx \\ &=\int 1\, dx-\int 3x^{-2}\, dx+\int 5x^{-4}\, dx &&\text{Rule 5}\\ &=\int 1\, dx-3\int x^{-2}\, dx+5\int x^{-4}\, dx &&\text{Rule 3}\\ &=x+C_1-3\left(\frac{x^{-1}}{-1}\right)+C_2+5\left(\frac{x^{-3}}{-3}\right)+C_3&&\text{Rules 2 and 4}\\ &=x+\frac{3}{x}-\frac{5}{3x^3}+C \\ \end{align*}


Therefore, the antiderivative of f(x)=\displaystyle\frac{x^4-3x^2+5}{x^4} is 


      \[F(x)=x+\frac{3}{x}-\frac{5}{3x^3}+C,\]


 where C is a real number constant. That is, the derivative of 


      \[F(x)=x+\frac{3}{x}-\frac{5}{3x^3}+C\]


 is f(x)=\displaystyle\frac{x^4-3x^2+5}{x^4}.


An integral can come in many different forms, and often, we need to rewrite the integral so that it is in a form that can be integrated. One such strategy of rewriting integrals is integration by parts.


Integration by Parts


We have discussed how to integrate integrals using only the basic integration rules. But what happens when we have a function that cannot be integrated this way? Consider the function 


      \[f(x)=x\sqrt{x-5}.\]


 We can write f(x) as the product of two functions, 


      \[u(x)=x\,\,\text{and}\,\,v(x)=\sqrt{x-5}.\]


 None of our basic integration rules tell us how to integrate products of functions so we need another method. This is where integration by parts comes in. Larson and Edwards give us the following theorem [15].


Theorem I.8


If u and v are functions of x and have continuous derivatives, then 


      \[\int u\,dv=uv-\int v\,du.\]




Proof.

Let u and v be functions of x with continuous derivatives.


      \begin{align*} \frac{d}{dx}[uv]&=u\frac{dv}{dx}+v\frac{du}{dx} &&\text{Product Rule}\\ \int\frac{d}{dx}[uv]\,dx&=\int u\frac{dv}{dx}\,dx+\int v\frac{du}{dx}\,dx &&\text{Integration Rule 5}\\ uv&=\int u\,dv+\int v\,du &&\text{dx's cancel}\\ \int u\,dv&=uv-\int v\,du &&\text{Rewrite}\\ \end{align*}


The process is to break the function we are trying to integrate into two separate parts, u and \frac{dv}{dx}. Then, differentiate u to get du and integrate \frac{dv}{dx} to get v. Finally, subtract the integral of v\,du from the product uv.


Let’s now revisit our function 


      \[f(x)=x\sqrt{x-5}\]


 and integrate it. This is a problem I completed for Calculus II that comes from Larson and Edwards [15].




Example 9

Integrate f(x)=x\sqrt{x-5}.


Solution

Let u=x and \frac{dv}{dx}=(x-5)^{\frac{1}{2}}. It follows then that


      \begin{align*} \frac{du}{dx}&=1\\ du&=1\, dx \end{align*}


      \begin{align*} \int dv&=\int (x-5)^{\frac{1}{2}}\, dx\\ v&=\frac{2(x-5)^{\frac{3}{2}}}{3} \end{align*}


So,


      \begin{align*} \int x\sqrt{x-5}\,dx&= x\left(\frac{2(x-5)^{\frac{3}{2}}}{3}\right)-\int \frac{2(x-5)^{\frac{3}{2}}}{3}(1) \,dx &&\text{Theorem \ref{Integration by Parts}}\\ &=\frac{2x(x-5)^{\frac{3}{2}}}{3}-\frac{2}{3}\int (x-5)^{\frac{3}{2}} \, dx \\ &=\frac{2x(x-5)^{\frac{3}{2}}}{3}-\frac{2}{3}\left(\frac{(x-5)^{\frac{5}{2}}}{\frac{5}{2}}\right)+C \\ &=\frac{2x(x-5)^{\frac{3}{2}}}{3}-\frac{4(x-5)^{\frac{5}{2}}}{15} +C \end{align*}


Therefore, the antiderivative of f(x)=x\sqrt{x-5} is 


      \[F(x)=\frac{2x(x-5)^{\frac{3}{2}}}{3}-\frac{4(x-5)^{\frac{5}{2}}}{15}+C.\]




Keep in mind that there are many different methods and formulas for integration so integration by parts is not guaranteed to be the best method every time. But when faced with an integral that involves multiplying or dividing functions, it is a good method to try.


Another method that we have for rewriting integrals is trigonometric substitution.


Trigonometric Substitution


Trigonometric substitution is a very helpful method when integrating radicals. However, it is limited in that it only works with three general forms. The following summary of the forms and substitutions is from Larson and Edwards [15].


 




 


In order to understand why these substitutions work, we need a couple of Pythagorean identities [15]:


    

  1. \sin^2\theta +\cos^2\theta =1
    2. 

  2. \sec^2\theta-\tan^2\theta=1
    3. 



All that we need to do is substitute u in and simplify using these Pythagorean identities. For example, the second form of trigonometric substitution looks like this:


      \begin{align*} \sqrt{a^2+u^2}&=\sqrt{a^2+a^2\tan^2\theta} &&\text{Subsitute $u$.}\\ &=\sqrt{a^2(1+\tan^2\theta)} &&\text{Factor out $a^2$.}\\ &=\sqrt{a^2\sec^2\theta} &&\text{Pythagorean identity 2.}\\ &=a\sec\theta &&\text{Simplify} \end{align*}


The other two follow a similar line of reasoning. Now, we will go through a problem I completed for Calculus II from Larson and Edwards [15].




Example 10


Find \displaystyle\int \displaystyle\frac{1}{\sqrt{16-x^2}} \,dx.


‌


Solution

Let u=x and a=4. By trigonometric substitution,


      \begin{align*} x&=4\sin\theta\\ \frac{dx}{d\theta}&=4\cos\theta \\ dx&=4\cos\theta \, d\theta\\ \text{}\\ \sqrt{16-x^2}&=4\cos\theta \end{align*}


Additionally,


      \begin{align*} x&=4\sin\theta\\ \frac{x}{4}&=\sin\theta\\ \theta&=\sin^{-1}\left(\frac{x}{4}\right) \end{align*}


Substituting everything in, we have


      \begin{align*} \int\frac{1}{\sqrt{16-x^2}} \,dx &=\int\frac{4\cos\theta}{4\cos\theta}\,d\theta \\ &=\int 1\,d\theta\\ &=\theta +C \\ &=\sin^{-1}\left(\frac{x}{4}\right)+C \end{align*}


Therefore, the antiderivative of \displaystyle\int \displaystyle\frac{1}{\sqrt{16-x^2}} \,dx is 


      \[F(x)=\sin^{-1}\left(\frac{x}{4}\right)+C.\]




What we have been discussing so far is the indefinite integral. But there is another type of integral known as the definite integral.


Definite Integrals


The definite integral is denoted as 


      \[\int_a^b\]


 This simply means that we are integrating over a closed interval [a,b]. We can evaluate such integrals using the following theorem given by Larson and Edwards [15].


Theorem I.9 (The Fundamental Theorem of Calculus)


If a function f is continuous on the closed interval [a,b] and F is an antiderivative of f on the interval [a,b], then 


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




We are now going to look at some of the many real-world applications of derivatives and integrals. Additionally, we will work through a real-world problem involving derivatives and another involving integrals, which demonstrates the Fundamental Theorem of Calculus.





					
							

						

							

								
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