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Integral Calculus – Definition, Techniques, and Application
Integral calculus is an important branch of calculus where we explore and understand the concepts behind integrals, their properties, as well as their applications. A lot of STEM specializations depend on integral calculus – including physics, engineering, biology, finance, and even sports analysis. This is why it’s important to learn about the components and concepts covered by integral calculus.
The integral calculus shows us how we can reverse the process of differential calculus. Indefinite integrals show us how we can use $\boldsymbol{f^{\prime}(x)}$ to determine the expression for $\boldsymbol{f(x)}$. Definite integrals return the numerical value when the antiderivative is evaluated at the limits of a given interval.
Our discussion will focus on the important components of integral calculus. We’ll show you the relationship shared between differential and integral calculus through the fundamental theorem of calculus. In this article, we’ll also show the difference between indefinite and definite integrals.
What is an integral in calculus?
Integral calculus is an equally essential part of calculus and is closely related to differential calculus. Integral calculus helps us in calculating and estimating areas under the curve. It also shows us how we can reverse the process of differential calculus to observe a related function or expression.
The integral of a function shows us what the area under the curve represents for an original function. The area can either be a fixed value or a new function – this will depend on whether we’re looking for the function’s definite or indefinite integral.
For now, let’s understand how the area under a function’s curve is related to the function’s integral.
In the past, we’ve learned how to estimate the area under the curve using the Riemann sum and other numerical methods. Integral calculus gives us the tool to approximate the area’s value as well as calculate its actual values whenever possible.
\begin{aligned}\text{Area} &= \int_{a}^{b} f(x)\phantom{x}dx\\&= F(b) – F(a)\end{aligned}
Breaking down the equations shown above, we have the following:
- The symbol, $\int$, represents the integral symbol.
- The area represents the definite integral of $f(x)$ evaluated at $x =a$ and $x =b$.
- The function, $F(x)$, represents the function’s indefinite integral or antiderivative.
Don’t worry, we’ll discuss these terminologies in the next sections. By the end of our discussion, we’ll be confident enough to differentiate indefinite and definite integrals. We’ll also have an idea of how we can calculate the function’s indefinite and definite integrals.
Indefinite integral calculus definition
The expression $\int f(x) \phantom{x}dx$ represents the indefinite integral or antiderivative of $f(x)$. The integral symbol, $\int$, indicates that we need to think of a function that when differentiated, returns $f(x)$. Let’s say $F(x)$ represents the antiderivative of the function, $f(x)$, for it to satisfy the definition of indefinite integral is as shown below:
\begin{aligned}F^{\prime}(x) &= f(x)\\\int f(x)\phantom{x} dx &= F(x) + C\end{aligned}
When given the integral expression, we call $\boldsymbol{f(x)}$ as the integrand. Meanwhile, $C$ represents the arbitrary constant that can be added to the antiderivative. For example, the five functions will all have $2x$ as their derivative. This is why we need to always include $C$ as a reminder that any function, $F(x)$, followed by $C$, will still have a derivative of $f(x)$. This means that $x^2 + C$ is the antiderivative of $2x$.
\begin{aligned}\dfrac{d}{dx} (x^2 +C) &= 2x\\\int 2x \phantom{x}dx &= x^2 +C\end{aligned}
These two equations clearly show the relationship between differential and integral calculus.
Want to learn more about indefinite integral and antiderivatives? Review the article we wrote here. For now, let’s go ahead and review what we know of definite integrals.
Definite integral calculus definition
The definite integral of a function will return a numerical value. This represents the area below the curve bounded by the vertical lines, $x =a$ and $x =b$. From the fundamental theorem of calculus, we’ve learned that when $F(x)$ is the antiderivative of $f(x)$, the definite integral can be defined as shown below:
\begin{aligned}\int_{a}^{b} f(x)\phantom{x}dx &= F(x)|_{a}^{b}\\ &=F(b) – F(a)\end{aligned}
This means that the value of $\int_{a}^{b} f(x)\phantom{x} dx$ is simply equivalent to the difference between antiderivative when evaluated at the lower and upper limits, $x =a$ and $x =b$.
These two important definitions also show us the two main applications of the integral calculus:
- We now have a method to find the expression for $f(x)$ when we’re only given $f^{\prime}(x)$. This concept has a wide range of applications in finance, physics, biology, and more.
- We’ve also shown how definite integrals will help us in calculating the area found under the curve of a function. This will come in handy when we need to approximate areas between two curves.
Seeing how important definite and indefinite integrals are in math and other fields, it’s essential that we know how to determine the function and values of antiderivatives. In the next section, we’ll recall the important integral properties and antiderivative formulas we’ll need to evaluate integrals completely.
How to do an integral in calculus?
To find the integral of a function, $f(x)$, we simply think of a function that when differentiated will return $f(x)$. We can think of derivative rules that could have been applied then work backward to find the antiderivative of a function.
This process, however, will be tedious, so we’ve summarized the essential antiderivative formula and integral properties that you’ll be using most of the time. Below is the table of the most common antiderivative formulas. Familiarizing yourself with these formulas will eventually make finding integrals more efficient.
Antiderivative Formulas | |
\begin{aligned}\int k\cdot f(x)\phantom{x}dx &= k \int f(x)\phantom{x}dx \end{aligned} | \begin{aligned}\int e^x \phantom{x}dx &= e^x + C \end{aligned} |
\begin{aligned} \int [f(x) \pm g(x)] \phantom{x}dx &= \int f(x)\phantom{x} dx \pm \int g(x)\phantom{x} dx \end{aligned} | \begin{aligned}\int \cos x \phantom{x}dx &= \sin x + C \end{aligned} |
\begin{aligned}\int k \phantom{x}dx &= kx + C \end{aligned} | \begin{aligned}\int \sin x \phantom{x}dx &= -\cos x + C \end{aligned} |
\begin{aligned}\int x^n \phantom{x}dx &= \dfrac{x^{n + 1}}{n + 1} + C \end{aligned} | \begin{aligned}\int \sec^2 x \phantom{x}dx &= \tan x + C \end{aligned} |
\begin{aligned}\int \dfrac{1}{x} \phantom{x}dx &= \ln|x| + C \end{aligned} | \begin{aligned}\int \csc x \cot x \phantom{x}dx &= -\csc x + C \end{aligned} |
\begin{aligned}\int \dfrac{1}{\sqrt{1 – x^2}} \phantom{x}dx &= \sin^{-1}x + C \end{aligned} | |
\begin{aligned}\int \dfrac{1}{1 + x^2} \phantom{x}dx &= \tan^{-1}x + C \end{aligned} | |
\begin{aligned}\int \dfrac{1}{x\sqrt{x^2 – 1}} \phantom{x}dx &= \sec^{-1}|x| + C \end{aligned} |
Here are some helpful pointers to keep in mind when evaluating indefinite integrals:
- Factor out constants whenever possible.
- Distribute the integral operation when functions being integrated contain multiple terms.
- Identify the antiderivative formulas that we might have to use to simplify the indefinite integral’s expression.
Example of Evaluating Indefinite Integrals |
\begin{aligned}\int (4x^3 – 8x^2) \phantom{x}dx &= \int 4x^3 \phantom{x}dx – \int 8x^2 \phantom{x}dx,\phantom{x}\color{Teal} \int [f(x) -g(x)]\phantom{x}dx =\int f(x) \phantom{x}dx-\int g(x) \phantom{x}dx \\&=4\int x^3 \phantom{x}dx – 8\int x^2 \phantom{x}dx, \phantom{x}\color{Teal} \int kf(x) \phantom{x}dx =k\int f(x) \phantom{x}dx \\&= 4\left(\dfrac{x^{3 + 1}}{3 + 1} \right ) – 8\left(\dfrac{x^{2 + 1}}{2 + 1} \right ) + C,\phantom{c}\color{Teal}\int x^n \phantom{x}dx =\dfrac{x^{n + 1}}{n +1} +C\\&= x^4 – \dfrac{8x^3}{3} + C \end{aligned} |
Apply a similar process when evaluating definite integrals, but once we have the antiderivative, $F(x) +C$:
- Disregard the arbitrary constant, $C$.
- Evaluate $F(x)$ at $x = a$ and $x = b$ then subtract the two values to find the value of $\int_{a}^{b} f(x)\phantom{x}dx$.
Properties of Definite Integrals | |
Sum or Difference | $\int_{a}^{b} [f(x) \pm g(x)]\phantom{x}dx = \int_{a}^{b} f(x) \phantom{x}dx \pm \int_{a}^{b} g(x) \phantom{x}dx $ |
Constant Multiple | $\int_{a}^{b} [k\cdot f(x)]\phantom{x}dx = k\int_{a}^{b} f(x) \phantom{x}dx$ |
Reverse Interval | $\int_{a}^{b} f(x)\phantom{x}dx = -\int_{b}^{a} f(x) \phantom{x}dx$ |
Zero-length Interval | $\int_{a}^{a} f(x)\phantom{x}dx = 0$ |
Combining Intervals | $\int_{a}^{b} f(x)\phantom{x}dx + \int_{b}^{c} f(x)\phantom{x}dx = \int_{a}^{c} f(x)\phantom{x}dx$ |
There are instances when we’ll have to apply some definite integral properties to manipulate the expression. The table above shows some essential definite integral properties to remember.
Here’s an example of evaluating definite integrals. When finding the value of $\int_{0}^{4} (2x^2 –x)\phantom{x} dx$, we first focus on finding the antiderivative of $2x^2 – x$. Once we have the indefinite integral, we’ll simply evaluate the expression at $x = 4$ and $x =0$ then subtract the resulting values.
Example of Evaluating Definite Integrals |
\begin{aligned}\int (2x^2 – x) \phantom{x}dx &= \int 2x^2 \phantom{x}dx – \int x \phantom{x}dx,\phantom{x}\color{Teal} \int [f(x) -g(x)]\phantom{x}dx =\int f(x) \phantom{x}dx-\int_{a}^{b} g(x) \phantom{x}dx \\&= 2\int x^2 \phantom{x}dx – \int x \phantom{x}dx, \phantom{x}\color{Teal} \int_{a}^{b} kf(x) \phantom{x}dx =k\int_{a}^{b} f(x) \phantom{x}dx \\&= 2\left(\dfrac{x^{2 + 1}}{2 + 1} \right ) – \left(\dfrac{x^{1 + 1}}{1 + 1} \right ) + C,\phantom{c}\color{Teal}\int x^n \phantom{x}dx =\dfrac{x^{n + 1}}{n +1} +C\\&= \dfrac{2x^3}{3} – \dfrac{x^2}{2} + C\\\\\int_{0}^{4} (2x^2 – x) \phantom{x}dx &= \dfrac{2x^3}{3} – \dfrac{x^2}{2}|_{0}^{4}\\&=\left[\left(\dfrac{2\cdot 4^3}{3} -\dfrac{4^2}{2}\right ) -\left(\dfrac{2\cdot 0^3}{3} -\dfrac{0^2}{2}\right ) \right ]\\&= \dfrac{104}{3}\end{aligned} |
What are some applications of integral calculus?
Now that we know how to evaluate indefinite and definite integrals, it’s time that we learn how to apply these techniques to solve word problems. One of the most common types of problems that uses integral calculus is the initial-value problem.
These are examples of word problems that require our knowledge of integral calculus for us to solve them:
- Finding the initial velocity of an object given the function representing the object’s acceleration.
- Calculating the average value of a function between a given interval.
- Estimating the amount of work done by an object over a period of time.
Let’s say we have $\dfrac{d}{dx}f(x) = 2x^2$ and $f(0) = 6$, we can use integral calculus to find the expression of $f(x)$. Since we have the derivative of $f(x)$, apply antiderivative formulas to find the indefinite integral of $f(x)$.
- Since $f^{\prime}(x) = 2x^2$, we have $\int f^{\prime}(x) \phantom{x}dx = f(x) +C$.
- Factor out $2$ from the integral expression using the property, $\int kf(x)\phantom{x}dx =k\int f(x)\phantom{x}dx$.
- Use the power rule,$\int x^n \phantom{x}dx = \dfrac{x^{n+1}}{n + 1} + C$, to find the indefinite integral of $f^{\prime}(x)$.
\begin{aligned}\int 2x^2 \phantom{x}dx &= 2\int x^2\phantom{x}dx\\&= 2\left(\dfrac{x^{2 +1}}{2 + 1} \right ) + C\\&= \dfrac{2x^3}{3} + C\end{aligned}
This means that $f(x) = \dfrac{2x^3}{3} + C$. Now, to find the value of $C$, use the fact that $f(x) = 6$ when $x =0 $.
\begin{aligned}f(x) &= \dfrac{2x^3}{3} + C\\6 &= \dfrac{2 \cdot0^3}{3} + C\\C &= 6\end{aligned}
Hence, $f(x) = \dfrac{2x^3}{3} + 6$. This is an actual example of an initial value-problem. This shows how we can find a function’s expression given its derivative and initial value.
We’ve now shown you the different properties in integral calculus and we’ve also shown you how to use these properties to find other expressions and values. Don’t worry, we have prepared enough problems for you to practice what you’ve just learned.
Example 1
If $f^{\prime}(x) = x^3 – 2x^2 + 4$, what is $f(x)$’s expression? Verify your answer by differentiating the resulting expression for $f(x)$.
Solution
Using the definition for antiderivatives, $\int f^{\prime}(x) \phantom{x} dx = f(x)$. This means that we can find $f(x)$ by integrating $x^3 – 2x^2 + 4$ using the different antiderivative formulas.
- Apply the sum or difference property for indefinite integrals, $\int [f(x) \pm g(x)] \phantom{x}dx = \int f(x) \phantom{x}dx \pm \int g(x) \phantom{x}dx $.
- Factor out $2$ from the second term using the property, $\int kf(x)\phantom{x} dx = k\int f(x) \phantom{x}dx$.
- Apply the power and constant rules for integrals: $\int x^n\phantom{x}dx = \dfrac{x^{n + 1}}{n + 1} +C$ and $\int k \phantom{x}dx = kx + C$.
\begin{aligned}\int (x^3 – 2x^2 + 4)\phantom{x}dx &= \int x^3\phantom{x}dx- \int 2x^2\phantom{x}dx + \int 4\phantom{x}dx\\&= \int x^3\phantom{x}dx- 2\int x^2\phantom{x}dx + \int 4\phantom{x}dx\\&=\left(\dfrac{x^{3 + 1}}{3 + 1}\right ) -2\left(\dfrac{x^{2 + 1}}{2 + 1}\right ) + \int 4\phantom{x}dx\\ &=\dfrac{x^4}{4} – \dfrac{2x^3}{3} + 4x + C \end{aligned}
This means that $f(x)$ is equal to $\dfrac{x^4}{4} – \dfrac{2x^3}{3} + 4x +C$, where $C$ is an arbitrary constant. To verify that our expression for $f(x)$ is indeed correct, let’s differentiate $\dfrac{d}{dx} f(x) = \dfrac{x^4}{4} – \dfrac{2x^3}{3} + 4x + C$.
\begin{aligned}f^{\prime}(x) &= \dfrac{d}{dx} f(x)\\&= \dfrac{d}{dx}\left(\dfrac{x^4}{4} – \dfrac{2x^3}{3} + 4x + C \right)\\&= \dfrac{d}{dx}\dfrac{x^4}{4} – \dfrac{d}{dx}\dfrac{2x^3}{3} + \dfrac{d}{dx}4x + \dfrac{d}{dx}C,\phantom{x}\color{Teal}\text{Sum or Difference Property}\\&= \dfrac{1}{4}\dfrac{d}{dx}x^4 – \dfrac{2}{3}\dfrac{d}{dx}x^3 + 4\dfrac{d}{dx}x + \dfrac{d}{dx}C,\phantom{x}\color{Teal}\text{Constant Multiple Property}\\&= \dfrac{1}{4}(4x^3) -\dfrac{2}{3}(3x^2) + 4(1) + \dfrac{d}{dx}C,\phantom{x}\color{Teal}\text{Power Rule}\\&= x^3 -2x^2 +4 + 0,\phantom{x}\color{Teal}\text{Constant Rule}\end{aligned}
From this, you can see that the properties used are similar but in reverse order. This shows how an indefinite integral is a result of reversing the process for derivatives.
Example 2
Evaluate the indefinite integral, $\int \dfrac{x^6 – 2x^5 + 1}{x^3}\phantom{x} dx$. Use the result to evaluate the definite integral, $\int_{1}^{4} \dfrac{x^6 – 2x^5 + 1}{x^3}\phantom{x} dx$.
Solution
Since integrals have no quotient rules as opposed to derivatives, we’ll rewrite $\dfrac{x^6 – 2x^5 + 1}{x^3}$ by dividing the numerator by denominator.
\begin{aligned}\int \dfrac{x^6 -2x^5 + 1}{x^3}\phantom{x} dx &= \int \left(\dfrac{x^6}{x^3} – \dfrac{2x^5}{x^3} + \dfrac{1}{x^3}\right )\phantom{x} dx\\&= \int \left(x^3 – 2x^2 + \dfrac{1}{x^3}\right)\phantom{x}dx\end{aligned}
Apply the antiderivative formulas we’ve just learned to evaluate the indefinite integral.
- Distribute the integral expression using the sum or difference property of integrals, $\int [f(x) \pm g(x)]\phantom{x}dx = \int f(x) \phantom{x}dx \pm \int g(x) \phantom{x}dx$.
- Factor out $2$ from the second term.
- Use the power rule, $\int x^n \phantom{x}dx = \dfrac{x^{n + 1}}{n + 1}\phantom{x} dx$, to integrate each term.
\begin{aligned}\int \left(x^3 – 2x^2 + \dfrac{1}{x^3}\right)\phantom{x}dx &=\int x^3 \phantom{x}dx- \int 2x^2\phantom{x}dx + \int\dfrac{1}{x^3}\phantom{x}dx\\&= \int x^3 \phantom{x}dx- \int 2x^2\phantom{x}dx + \int x^{-3}\phantom{x}dx\\&= \int x^3 \phantom{x}dx- 2\int x^2\phantom{x}dx + \int x^{-3}\phantom{x}dx \\&= \left( \dfrac{x^{3 + 1}}{3 + 1}\right ) – 2\left( \dfrac{x^{2 + 1}}{2 + 1}\right )+ \left( \dfrac{x^{-3 + 1}}{-3 + 1}\right ) + C\\&= \dfrac{x^4}{4} – \dfrac{2x^3}{3} – \dfrac{x^{-2}}{2} + C\\&= \dfrac{x^4}{4} – \dfrac{2x^3}{3} – \dfrac{1}{2x^2} + C\end{aligned}
Hence, we have $\int \dfrac{x^6 – 2x^5 + 1}{x^3}\phantom{x} dx = \dfrac{x^4}{4} – \dfrac{2x^3}{3} – \dfrac{1}{2x^2} + C $.
Evaluate the definite integral by evaluating the antiderivative at the lower and upper limits. Disregard $C$ for the definite integrals since we’re looking for a specific numerical value.
\begin{aligned}\int_{1}^{4} \dfrac{x^6 – 2x^5 + 1}{x^3}\phantom{x} dx &= \dfrac{x^4}{4} – \dfrac{2x^3}{3} – \dfrac{1}{2x^2} |_{1}^{4}\\&= \left[\left(\dfrac{8^4}{4}-\dfrac{2\cdot8^3}{3} – \dfrac{1}{2\cdot8^2} \right ) -\left(\dfrac{1^4}{4}-\dfrac{2\cdot1^3}{3} – \dfrac{1}{2\cdot1^2} \right )\right ]\\&= \left(\dfrac{2045}{96} -\dfrac{11}{12}\right )\\&= \dfrac{711}{32}\end{aligned}
This means that $\int_{1}^{4} \dfrac{x^6 – 2x^5 + 1}{x^3}\phantom{x} dx = \dfrac{711}{32}$.
Example 3
Alexa was driving her car at the rate of $40 \text{ km/h}$ when she pressed on the brakes. Alexa’s car begins decelerating at a constant rate of $8 \text{ km/h}^2$. How many seconds have passed before Alexa’s car stops?
Hint: The acceleration function, $a(t)$, is simply the derivative of the velocity function, $v(t)$.
Solution
The velocity of Alexa’s car at first (or at time, $t =0$), is equal to $40 \text{ km/h}$, so $v(0) = 40$. Find the expression for $v(t)$ by integrating $a(t)$.
\begin{aligned}v(t) &= \int a(t) \phantom{x} dt\end{aligned}
Recall that the car is decelerating or slowing down, so $a(t)$ is negative. Apply the constant rule, $\int k \phantom{x}dx = kx + C$, to find the expression for $v(t)$.
\begin{aligned}v(t) &= \int -8\phantom{x} dt\\&= -8t + C\end{aligned}
From this, we can see that this is an initial-value problem with us having the following values: $v(t) = -8t$ and $v(0) = 40$.
Solve for $C$ using these values to complete the expression for $v(t)$.
\begin{aligned}v(0) &= 40\\-8(0) + C &= 40\\C &= 48\end{aligned}
Hence, we have $v(t) = -8t + 48$. When Alexa’s car comes to a halt, $v(t) = 0$. Equate $v(t)$ to $0$ and solve for $t$ to find the time it takes for the car to stop.
\begin{aligned}-8t + 48 &= 0\\-8t &= -48\\t &= 6\end{aligned}
It takes $6$ seconds for Alexa’s car comes to a stop.
Practice Questions
1. If $f^{\prime}(x) = -4x^5 + 8x^3 – x + 1$, what is $f(x)$’s expression? Verify your answer by differentiating the resulting expression for $f(x)$.
2. Evaluate each of the following indefinite integrals:
a. $\int \dfrac{-2x^8 – 4x^6 + x^2}{x^4}\phantom{x} dx$
b. $\int \dfrac{4}{x^2 + 1}\phantom{x} dx$
c. $\int (\cos x + \tan x \cos x)\phantom{x}dx$
3. Evaluate each of the following definite integrals:
a. $\int_{0}^{5} (5x^2 – 1)\phantom{x} dx$
b. $\int_{1}^{4} \dfrac{x^6 – 2x^5 + 1}{x^2}\phantom{x} dx$
c. $\int_{0}^{\pi} (\cos x – \sin x)\phantom{x}dx$
Answer Key
1. $f(x) = -\dfrac{2x^6}{3}+ 2x^4- \dfrac{x^2}{2}+x +C$
$\begin{aligned}f^{\prime}(x) &= \dfrac{d}{dx} f(x)\\&= \dfrac{d}{dx}\left(-\dfrac{2x^6}{3}+ 2x^4- \dfrac{x^2}{2}+x +C\right)\\&= -\dfrac{2}{3}(6x^5) + 2(4x^3) – \dfrac{1}{2}(2x) + 1 + 0\\&= -4x^5 + 8x^3 -x + 1\end{aligned}$
2.
a. $\int \dfrac{-2x^8 – 4x^6 + x^2}{x^4}\phantom{x} dx = -\dfrac{2x^5}{5}- \dfrac{4x^3}{3} -\dfrac{1}{x} +C$
b. $\int \dfrac{4}{x^2 + 1}\phantom{x} dx = 4\tan^{-1}x + C$
c. $\int (\cos x + \tan x \cos x)\phantom{x}dx = \sin x – \cos x +C$
3.
a. $\int_{0}^{5} (5x^2 – 1)\phantom{x} dx = \dfrac{610}{3}$
b. $\int_{1}^{4} \dfrac{x^6 – 2x^5 + 1}{x^2}\phantom{x} dx = \dfrac{1557}{20}$
c. $\int_{0}^{\pi} (\cos x – \sin x)\phantom{x}dx = -2$
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