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Introduction to Integration Calculus: Definitions, Formulas, & Examples

03.15.2022 • 9 min read

Rachel McLean

Subject Matter Expert

In this article, we’ll discuss the definition of integral calculus, indefinite vs definite integration, key integration rules, and three essential integration methods.

In This Article

  1. What is Integral Calculus?

  2. Standard Integration Rules and Theorems

  3. Indefinite vs Definite Integrals

  4. 3 Ways to Calculate Integrals

What is Integral Calculus?

You are probably already familiar with differentiation, which is the process used to calculate the instantaneous rate of change of a function. What is the difference between integration and differentiation? Well, you can think about integration as the reverse operation of differentiation. Together, differentiation and integration make up the essential operations of calculus and are related by the fundamental theorems of calculus.

Dr. Hannah Fry discusses the fundamental theorem of calculus:

When you integrate some function f(x)f(x), you find its antiderivative function, which is often denoted F(x)F(x). This function can ‌calculate the area underneath the curve of f(x)f(x).

The notation for integrating f(x)f(x) looks like this:

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

Here’s a guide for interpreting this integral notation:

  • What is \int?

The symbol \int is called the integral sign. This symbol indicates that we’re calculating the antiderivative function of f(x)f(x).

  • What is f(x)f(x)?

The function f(x)f(x) is called the integrand, and it’s the function we’re taking the integral of.

  • What is dxdx in calculus?

These letters represent the differential dxdx. The differential dxdx indicates that we’re integrating f(x)f(x) with respect to the variable xx.

  • What is F(x)F(x)?

F(x)F(x) is the antiderivative function that gives back f(x)f(x) when differentiated.

  • What is CC?

The capital letter CC represents a constant value called the constant of integration. We’ll talk more about what the constant of integration means later.

  • Why is F(x)F(x) called the antiderivative function?

When you take the derivative of F(x)F(x), you get back f(x)f(x) again. To help think about the relationship between a function ff and its antiderivative, you can ask the question, “What function F(x)F(x) has the derivative f(x)f(x)?” Their relationship can be thought of like this:

If f(x)dx=F(x)+C\int f(x)\,dx = F(x) + C, then F(x)=f(x)F’(x) = f(x)

Or, it might be easier to think about it like this:

f(x)dx=f(x)+C\int f’(x)\,dx = f(x) + C

For example,

Graph showing integration, differentiation, and relationship guaranteed by the First Fundamental Theorem of Calculus
Graphic showing integration, differentiation, and relationship guaranteed by the First Fundamental Theorem of Calculus

This relationship is guaranteed by the First Fundamental Theorem of Calculus:

F(x)=ddxaxf(t)dt=f(x)F'(x) = \frac{d}{dx} \int_a^x f(t)\,dt = f(x)

Because of this relationship, you might also hear integration referred to as anti-differentiation.

Dr. Tim Chartier discusses why we need antiderivatives to begin with:

Standard Integration Rules and Theorems

Assuming that ff and gg are continuous functions, here is a list of the most important integration rules and properties that you should know:

Sum Rule

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

Difference Rule

[f(x)g(x)]dx=f(x)dxg(x)dx\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx

Constant Multiplier Rule

kf(x)dx=kf(x)dx\int kf(x)\,dx = k\int f(x)\,dx for some constant kk

Power Rule

xndx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C for some real number nn

Constant Rule

adx=ax+C\int a\,dx = ax + C for some constant aa

Reciprocal Rules

1xdx=x1dx=lnx+C\int \frac{1}{x}\,dx = \int x^{-1}\,dx = \ln{|x|} + C

1ax+bdx=1aln(ax+b)+C\int\frac{1}{ax+b}\,dx = \frac{1}{a} \ln{(ax+b)} + C

Exponential and Logarithmic Function Rules

exdx=ex+C\int e^x\,dx = e^x + C

axdx=axln(a)+C\int a^x\,dx = \frac{a^x}{\ln{(a)}} + C, for any positive real number aa

ln(x)dx=xln(x)x+C\int \ln{(x)}\,dx = x\ln{(x)}-x + C

Trigonometric Function Rules

For these rules, assume that x is in radians.

sinxdx=cosx+C\int \sin{x}\,dx = -\cos{x} + C

cosxdx=sinx+C\int \cos{x}\,dx = \sin{x} + C

sec2xdx=tanx+C\int \sec ^2 x\,dx = \tan{x} + C

csc2xdx=cotx+C\int \csc ^2 x\,dx = -\cot{x} + C

secxtanxdx=secx+C\int \sec{x}\tan{x}\,dx = \sec{x} + C

cscxcotxdx=cscx+C\int \csc{x}\cot{x}\,dx = -\csc{x} + C

dx1x2=sin1x+C\int \frac{dx}{\sqrt{1-x^2}}=\sin ^{-1}x + C

dx1x2=cos1x+C\int \frac{-dx}{\sqrt{1-x^2}}=\cos ^{-1}x + C

dx1+x2=tan1x+C\int \frac{dx}{1+x^2}=\tan ^{-1}x + C

sin(ax)dx=cos(ax)a+C\int \sin{(ax)}\,dx = \frac{-\cos{(ax)}}{a}+C for some real number aa

cos(ax)dx=sin(ax)a+C\int \cos{(ax)}\,dx = \frac{\sin{(ax)}}{a}+C for some real number aa

Absolute Value Rule

xdx=xx2+C\int |x|\,dx = \frac{x |x|}{2} + C

Here are a few short examples to practice these integration rules.

  • Practice the power rule:

x4dx=x55+C\int x^4\,dx = \frac{x^{5}}{5} + C

  • Practice the constant multiplier rule:

7x6dx=7x6dx=7x77=x7+C\int 7x^6\,dx = 7\int x^6\,dx = 7 \cdot \frac{x^7}{7} = x^7 + C

  • Practice the constant rule:

32dx=32x+C\int 32\,dx = 32x + C

  • Practice the reciprocal rules:

12x+17dx=12ln(2x+17)+C\int\frac{1}{2x+17}\,dx = \frac{1}{2} \ln{(2x+17)} + C

  • Practice the sum/difference rules:

(x+sin(x)3x)dx=xdx+sin(x)dx3xdx=x22cosx3xln(3)+C\int (x + sin{(x)} - 3^x)\,dx = \int x \,dx + \int \sin{(x)}\,dx - \int 3^x \,dx= \frac{x^2}{2} -\cos{x} - \frac{3^x}{\ln{(3)}} + C

Indefinite vs Definite Integrals

There are two types of integrals: indefinite and definite.

Dr. Hannah Fry talks more about indefinite and definite integrals:

The indefinite integral finds the general antiderivative function of f(x)f(x), while the definite integral finds the area under the curve of f(x)f(x) on a specific interval.

graphic  showing indefinite and definite integrals having different outputs

These types of integrals have different outputs. The definite integral outputs a unique number that represents the area enclosed by a function’s curve and the x-axis over some interval [a,b][a, b]. The indefinite integral outputs a function’s antiderivative function, accompanied by the constant of integration CC.

So, what exactly is the constant of integration CC? Let’s figure this out by looking at an example. Consider the function f(x)=5x4f(x) = 5x^4. Using the power rule, observe that (5x4)dx=x5+C=F(x)\int (5x^4) \,dx = x^5 + C = F(x).

Let’s try plugging some different CC values into the antiderivative function F(x)=x5+CF(x) = x^5 + C. When we take the derivative of the antiderivative function F(x)F(x), we should get our original function f(x)f(x) back again.

  • If CC= 4, then F(x)=x5+4F(x) = x^5+4, and its derivative is F(x)=5x4F’(x) = 5x^4

  • If CC= 0, then F(x)=x5F(x) = x^5, and its derivative is F(x)=5x4F’(x) = 5x^4

  • If CC= -150, then F(x)=x5150F(x) = x^5-150, and its derivative is F(x)=5x4F’(x) = 5x^4

Even though their last constant term is different, all of the above F(x)F(x) functions have the same derivative F(x)F’(x), and F(x)F’(x) is always equal to our original integrand function f(x)=5x4f(x) = 5x^4.

Graph of Integration and differentiation example

This is because the derivative of any constant is zero. Remember that for an integrand f(x)f(x), its antiderivative function answers the question, “What function F(x)F(x) has the derivative f(x)f(x)?” Any of the above F(x)F(x) functions will satisfy this question.

Since there are infinite constant values we can plug into CC, the constant of integration CC and the antiderivative function F(x)F(x) together represent an infinite family of functions. That’s why it’s called “indefinite” integration since there is not one unique antiderivative function.

By contrast, definite integrals give us a unique answer. We also don’t need to worry about the constant of integration anymore. To evaluate the definite integral of a function f(x)f(x) on the interval [a,b][a, b], we use this notation:

abf(x)dx=A\int_{a}^{b} f(x)\,dx = A
notation showing integral bounds or limits

The letters aa and bb are called integral bounds or limits. The letter aa represents the lower bound, while bb represents the upper bound. We can think of this notation as the area bounded by f(x)f(x), the x-axis, and the lines x=ax=a and x=bx=b.

Graph showing Second Fundamental Theorem of Calculus

To find the definite integral of a function on [a,b][a, b], we take the difference between the indefinite integral of the function evaluated at aa and the indefinite integral of the function evaluated at bb. This is called the Second Fundamental Theorem of Calculus.

This theorem clarifies the relationship between the integral and the antiderivative function:

abf(x)dx=F(x)ab=F(b)F(a)\int_{a}^{b} f(x)\,dx = F(x)\Big|_a^b = F(b) - F(a)

Here are the four steps for evaluating a definite integral:

  • Step 1 - Find the indefinite integral F(x)F(x) using the integral rules.

  • Step 2 - Find F(b)F(b) by plugging bb into F(x)F(x).

  • Step 3 - Find F(a)F(a) by plugging aa into F(x)F(x).

  • Step 4 - Take the difference F(b)F(a)F(b) - F(a). Since we subtract these values, the constant of integration CC cancels out, which is why we can ignore it.

For example, let’s evaluate f(x)=x+1f(x) = x + 1 on [2,4][2, 4]. The notation for this looks like 24(x+1)dx\int_{2}^{4} (x + 1)\,dx.

  • Step 1 - Using the power and constant rules, (x+1)dx=x22+x\int( x + 1)\,dx = \frac{x^2}{2} + x.

  • Step 2 - F(4)=422+4=12F(4) = \frac{4^2}{2} + 4= 12

  • Step 3 - F(2)=222+2=4F(2) = \frac{2^2}{2} + 2 = 4

  • Step 4 - F(4)F(2)=124=8F(4) - F(2) = 12 - 4 = 8

This value represents the area under the curve of f(x)f(x) on [2,4][2, 4].

graph showing value of the area under the curve

3 Ways to Calculate Integrals

Below, we’ll discuss three primary techniques for evaluating more complex integrals.

1. U-Substitution

U-Substitution reverses the chain rule for derivatives and is used to integrate composite functions. We need to rewrite our integral in terms of uu and dudu, so that it looks like this:

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)\,dx = \int f(u)\,du

Here are the four steps to integrating with u-substitution:

  1. Pick uu, the "inside" part of the chain rule.

  2. Differentiate uu to find dudu. If necessary, rearrange the problem algebraically so that dudu perfectly matches what’s left inside the integral.

  3. Substitute uu and dudu into the integrand and integrate.

  4. Substitute the original values back into the resulting function and add the constant of integration to your final answer.


Let’s evaluate 10x(5x2+1)2dx\int \frac{10x}{(5x^2+1)^2}\,dx. Let u=5x2+1u = 5x^2 + 1. Then, du=10xdxdu = 10x\,dx. Now, we’ll substitute uu and dudu into the integral.

f(g(x))g(x)dx=f(u)du\int f(g(x))g’(x)\,dx = \int f(u)\,du
10x(5x2+1)2dx=duu2\int \frac{10x}{(5x^2+1)^2}\,dx = \int \frac{du}{u^2}
=u2du= \int u^{-2}\,du
=u11 = \frac{u^{-1}}{-1}
=15x2+1+C= \frac{-1}{5x^2+1} + C

2. Integration by Parts

Integration by parts uses this formula to integrate a product of functions :

udv=uvvdu\int udv = uv - \int v\,du

We must choose one function in the integrand to represent uu and the other to represent dvdv.

Here are four steps to integrate with integration by parts:

  1. Separate the integrand into a product of functions by choosing uu and dvdv.

  2. Differentiate uu to find dudu and integrate dvdv to find vv.

  3. Plug uu, vv, and dudu into the integration by parts formula.

  4. Solve and simplify.


Let’s evaluate xsin(x)dx\int x \sin(x)\,dx. We’ll let dv=sin(x)dxdv = \sin{(x)}\,dx since the integral of this function is easy to find . Then, u=xu = x since that’s what is left over. Now, we need to differentiate uu to find dudu and integrate dvdv to find vv.

Using the power rule on u=xu = x and solving for dudu, we find that du=1dxdu = 1dx. Integrating dvdv using the trigonometry rules, we find that v=sin(x)=cos(x)v = \int \sin{(x)} = -\cos{(x)}. Now we can plug these values into our formula.

udv=uvvdu\int u\,dv = uv - \int v\,du
xsin(x)dx=x(cos(x)cos(x)dx\int x \sin(x)\,dx = x(-\cos{(x)} - \int -\cos{(x)}\,dx
=xcos(x)+cos(x)dx= -x\cos{(x)} + \int \cos{(x)}\,dx
=xcos(x)+sin(x)= -x\cos{(x)} + \sin{(x)}

3. Integration by Partial Fractions

Integration by partial fractions is used to integrate rational functions. This method is hard to understand without an example, so be sure to try the example exercise.

Here are nine steps to integrating with this method:

  1. Factor the denominator of the function.

  2. Decompose the function into a sum of its parts by assigning an unknown variable to each term of the denominator.

  3. Combine all terms into one by finding the common denominator, making sure you multiply each numerator appropriately.

  4. Multiply out the numerator.

  5. Set up an equation that equates the xx terms of the original function’s numerator with the xx terms in your new equation’s numerator.

  6. Set up a second equation that equates the constant terms of the original function’s numerator with the constant terms in your new numerator.

  7. Solve for the unknown variables.

  8. Plug your solved variables into Step 2.

  9. Integrate using the reciprocal rules.


Let’s evaluate x+4x2+x12dx\int \frac{x+4}{x^2+x-12}dx. First, we need to factor the denominator.

x+4x2+x12dx=x+4(x+4)(x3)dx\int \frac{x+4}{x^2+x-12}\,dx = \int \frac{x+4}{(x+4)(x-3)}\,dx

Now we can do steps 2 - 4.

x+4(x+4)(x3)=Ax+4+Bx3\frac{x+4}{(x+4)(x-3)} = \frac{A}{x+4} + \frac{B}{x-3}
=A(x3)+B(x+4)(x+4)(x3)=\frac{A(x-3) + B(x+4)}{(x+4)(x-3)}

Now, we can solve for A and B with steps 5 - 7.

Following those steps, our first equation is Ax+Bx=xAx + Bx = x, which simplifies to A+B=1A+B=1. Our second equation is 3A+4B=4-3A+4B=4. Solving this system of equations, we find that A=0A = 0 and B=1B = 1.

We can now finish with steps 8 - 9:

x+4(x+4)(x3)dx=0x+4dx+1x3dx\int \frac{x+4}{(x+4)(x-3)}\,dx = \int \frac{0}{x+4}dx + \frac{1}{x-3}\,dx
=ln(x3)= \ln{(x-3)}

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