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Calculus

Integration in Math: Definition, How to Calculate It, and Examples

01.12.2022 • 10 min read

Rachel McLean

Subject Matter Expert

This article is a guide on how to integrate in calculus. Learn the definition of integration, how to evaluate integrals using 4 different methods, and practice with some examples.

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In This Article

  1. What Is Math Integration?

  2. Key Integration Formulas

  3. How to Calculate Integrals

  4. 4 Examples of Integration

What is Integration in Math?

What does integration mean in math? Integration is the reverse process of differentiation. Integrating some function f(x)f(x) outputs another function, F(x)F(x). When differentiated, this function F(x)F(x) gives back the original function f(x)f(x). For this reason, the indefinite integral is often called the antiderivative. Similarly, we also refer to integration as anti-differentiation.

To refresh your knowledge of integrals, read this beginner’s guide to integrals.

Outlier Graph Math Integration

Let’s review the notation for integration. The indefinite integral of a real-valued function f(x)f(x) with respect to a real variable x is written as:

f(x)dx=F(x)+C\int f(x)dx = F(x) + C
  • The symbol \int is called the integral sign.

  • The function f(x)f(x) is called the integrand. We want to find the antiderivative of this function.

  • The letters dxdx represent the differential dxdx. The differential dxdx indicates that we are integrating our function with respect to the variable x.

  • F(x)F(x) is the antiderivative function that results from taking the integral.

  • The capital letter C represents some constant value called the constant of integration.

Why is the constant of integration placed at the end of every indefinite integral?

For every f(x)f(x), we add C to its antiderivative F(x)F(x) in order to include all antiderivatives of f(x)f(x). Remember that the antiderivative provides a solution to the question, “What function F(x)F(x) gives back f(x)f(x) when differentiated?” If F(x)F(x) is the antiderivative of f(x)f(x), then their relationship looks like this:

F(x)=f(x)F’(x) = f(x)

An infinite number of functions fit this bill, since the derivative of any constant is simply zero. For a simple integration example, let f(x)=2xf(x) = 2x. Observe that (2x)dx=x2+C=F(x)\int (2x) dx = x^2 + C = F(x).

Now, notice the following:

  • If F(x)=x2+1F(x) = x^2+1, then F(x)=2xF’(x) = 2x

  • If F(x)=x2+2F(x) = x^2+2, then F(x)=2xF’(x)= 2x

  • If F(x)=x2+63F(x) = x^2+63, then F(x)=2xF’(x)= 2x

  • If F(x)=x2+100F(x) = x^2+100, then F(x)=2xF’(x) = 2x

No matter what constant we place at the end of the antiderivative function F(x)F(x), the derivative F(x)F’(x) is always the same; and F(x)F’(x) is always equal to the original function f(x)f(x).

Since there are an infinite number of functions F(x)F(x) whose derivative is f(x)f(x), the inclusion of C gives us a way to express the general form of the antiderivative F(x)F(x) by representing a family of infinitely many functions.

To explain more about antiderivatives, one of our instructors, Dr. Tim Chartier, talks more in depth about them:

Key Integration Formulas

Let’s review the most important rules of integration and some standard integrals. While there are many rules to memorize, don’t despair. Many follow a similar pattern. It’s worth your time to become familiar with them all.

In the following properties, assume that ff and gg are continuous functions.

Sum and Difference Rule

[f(x)±g(x)]dx=f(x)dx±g(x)dx\int [f(x) \pm g(x)]dx = \int f(x)dx \pm \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 k

Power Rule

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

Constant Rule

adx=ax+C\int adx = ax + C for some constant a

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^xdx = e^x + C

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

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 xdx = \tan{x} + C

csc2xdx=cotx+C\int \csc ^2 xdx = -\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 a

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

Absolute Value Rule

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

Dr. Tim Chartier explains these rules more here:

How to Calculate Integrals

We’ll learn how to integrate using 4 different methods. While integral calculators do exist, it’s crucial that you understand how to integrate on your own.

1. Simplification

Before evaluating any integral, determine if the integrand can be simplified. Simplifying the integrand using trigonometric identities, algebraic identities, and cancellation of similar terms can reveal clever ways to use the rules of integration above. For example, let’s calculate x32xx4dx\int \frac{x^3-2x}{x^4} dx. Notice that by simplifying and using the Sum Rule, we get:

x32xx4dx=x3x4dx+2xx4dx\int \frac{x^3-2x}{x^4} dx = \int \frac{x^3}{x^4} dx + \int \frac{-2x}{x^4} dx
=1xdx+2x3dx= \int \frac{1}{x} dx + \int \frac{-2}{x^3} dx

Now, we can apply the Reciprocal Rule, the Power Rule, and the Constant Multiplier Rule.

1xdx+2x3dx=lnx2x3dx\int \frac{1}{x} dx + \int \frac{-2}{x^3} dx = \ln{|x|} -2 \int x^{-3}dx
=lnx2(x22)= \ln{|x|} -2(\frac{x^{-2}}{-2})
=lnx+1x2+C=\ln{|x|} + \frac{1}{x^{2}}+C

So x32xx4dx=lnx+1x2+C\int \frac{x^3-2x}{x^4} dx =\ln{|x|} +\frac{1}{x^{2}}+C.

2. Integration by Parts

We use integration by parts for evaluating the integral of a product of functions.

Here’s the integration by parts formula:

udv=uvvdu\int udv = uv - \int vdu

Integration by parts involves choosing one function in your integrand to represent u and one function to represent dv.

Here are some simple steps:

1. Choose uu and dvdv to separate the given function into a product of functions.

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 where needed.

Generally, you choose uu to be the term that’s easiest to differentiate and choose dvdv to be the term that’s easiest to integrate.

We’ll do one example together. Say we want to evaluate xcos(3x)dx\int x \cos (3x) dx.

Step 1

Let’s choose dv=cos(3x)dxdv = \cos (3x)dx since we have an easy formula to integrate this function using the trigonometry rules above. Then, u=xu = x since that’s what’s left over.

Step 2

Differentiating u, we have dudx=d(x)dx=1\frac{du}{dx}=\frac{d(x)}{dx}=1. After multiplying both sides by the differential dxdx, we find that du=1dxdu = 1dx.

Now, let’s integrate dvdv to find vv. One of the trigonometric formulas above gives us an easy answer to this. So v=dv=cos(3x)dx=sin(3x)3v = \int dv = \int \cos (3x) dx = \frac{\sin{(3x)}}{3}.

Step 3

Using the integration by parts formula, we have:

udv=uvvdu\int udv = uv - \int vdu
xcos(3x)dx=xsin(3x)3sin(3x)3dx\int x \cos (3x) dx = x \cdot \frac{\sin{(3x)}}{3} - \int \frac{\sin{(3x)}}{3}dx
=xsin(3x)313sin(3x)dx= \frac{x\sin{(3x)}}{3} - \frac{1}{3} \cdot \int \sin{(3x)}dx
=xsin(3x)313cos(3x)3= \frac{x\sin{(3x)}}{3} - \frac{1}{3} \cdot \frac{-\cos{(3x)}}{3}
=xsin(3x)3+cos(3x)9+C= \frac{x\sin{(3x)}}{3} + \frac{\cos{(3x)}}{9}+C

So xcos(3x)dx=xsin(3x)3+cos(3x)9+C\int x \cos (3x) dx = \frac{x\sin{(3x)}}{3} + \frac{\cos{(3x)}}{9}+C.

3. U-Substitution

U-Substitution unravels the chain rule for derivatives. For this reason, u-substitution is also called the Reverse Chain Rule and is used to integrate composite functions.

If you can recognize both a function and its derivative inside the same integrand, that’s a good indication that you can use u-substitution. A function where you can use u-substitution would look like this:

f(g(x))g(x)dx\int f(g(x))g’(x)dx

We want to replace substitute u=g(x)u = g(x) and du=g(x)du = g’(x) into our integrand, so that it looks like this:

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

For our example, let’s evaluate (4x)e2x2+5dx\int (4x)e^{2x^2 + 5}dx. Notice that the derivative of 2x2+52x^2 + 5 is equal to 4x4x, and both of these equations are present in the integrand.

So we can substitute into our integrand u=2x2+5u = 2x^2+5 and du=4xdxdu = 4xdx. Then, we’ll substitute the real values back into the equation after integrating.

(4x)e2x2+5dx=eudu\int (4x)e^{2x^2 + 5}dx = \int e^u du
=eu+C=e^u + C
=e2x2+5+C=e^{2x^2+5}+C

So (4x)e2x2+5dx=e2x2+5+C\int (4x)e^{2x^2 + 5}dx = e^{2x^2+5}+C.

4. Integration by Partial Fractions

This method is useful for integrating rational functions. A rational function is a function that is written as the ratio of two polynomial functions. With this technique, we’ll decompose the integrand into several smaller rational functions.

The best way to explain this method is with an example. For our example, let’s calculate 2x2x2+9x+4dx\int \frac{2x}{2x^2+9x+4} dx. First, we need to factor the denominator. Then, we can split the integrand into two rational functions with unknown numerators and add the two new rational functions together. We want to solve for A and B.

2x2x2+9x+4=2x(2x+1)(x+4)\frac{2x}{2x^2+9x+4} = \frac{2x}{(2x+1)(x+4)}
=A2x+1+Bx+4= \frac{A}{2x+1} + \frac{B}{x+4}
=A(x+4)(2x+1)(x+4)+B(2x+1)(2x+1)(x+4)=\frac{A(x+4)}{(2x+1)(x+4)}+\frac{B(2x+1)}{(2x+1)(x+4)}
=Ax+4A+2Bx+B(2x+1)(x+4)=\frac{Ax+4A+2Bx+B}{(2x+1)(x+4)}

Now, we collect the x terms on one side of the numerator and collect the constant terms on the other side of the numerator. This allows us to mirror the structure of the original numerator of the integrand, 2x+02x+0.

2x(2x+1)(x+4)=(Ax+2Bx)+(4A+B)(2x+1)(x+4)\frac{2x}{(2x+1)(x+4)} = \frac{(Ax+2Bx)+(4A+B)}{(2x+1)(x+4)}

Now, we can solve for A and B by setting up two equations:

  • 2x=Ax+2Bx2x = Ax + 2Bx, which simplifies to 2=A+2B2 = A + 2B

This equation equates the x terms in the original integrand’s numerator with the x terms in the new rational function’s numerator.

  • 0=4A+B0 = 4A + B

This equation equates the constant terms in the original integrand’s numerator with the constant terms in the new rational function’s numerator.

We can now solve for A and B.

Solving the system of equations with 2x=Ax+2Bx2x = Ax + 2Bx and 0=4A+B0 = 4A + B gives us A=27A=\frac{-2}{7} and B=87B=\frac{8}{7}.

We’re finally at the last step. Since we’ve found A and B, we can plug these back into our function and calculate the integral. We’ll need to use the second Reciprocal Rule.

2x2x2+9x+4dx=(27(2x+1)+87(x+4))dx\int \frac{2x}{2x^2+9x+4} dx = \int (\frac{-2}{7(2x+1)}+\frac{8}{7(x+4)})dx
=2712x+1dx+871x+4dx=\frac{-2}{7}\int \frac{1}{2x+1}dx + \frac{8}{7}\int \frac{1}{x+4}dx
=2ln(2x+1)27+8ln(x+4)7+C=\frac{-2 \ln{(2x+1)}}{2 \cdot 7} + \frac{8 \ln{(x+4)}}{7} + C
=17ln(2x+1)+87ln(x+4)=-\frac{1}{7}\ln{(2x+1)} + \frac{8}{7} \ln{(x+4)}
=87ln(x+4)17ln(2x+1)+C=\frac{8}{7} \ln{(x+4)} -\frac{1}{7}\ln{(2x+1)} + C

So 2x2x2+9x+4dx=87ln(x+4)17ln(2x+1)+C\int \frac{2x}{2x^2+9x+4} dx =\frac{8}{7} \ln{(x+4)} -\frac{1}{7}\ln{(2x+1)} + C.

4 Examples of Integration

Let’s work through a few more integration examples together.

Example 1

Evaluate 6(x24x5x+1+sinx)dx\int 6(\frac{x^2-4x-5}{x+1} + \sin{x})dx.

Solution: First, let’s break this integrand into two separate functions using the Sum Rule and apply the Constant Multiplier Rule. Then, we’ll factor the numerator of the rational expression and use the trigonometry formulas. Finally, we'll cancel similar terms and apply the Sum Rule and Power Rule:

6(x24x5x+1+sinx)dx=6x24x5x+1dx+6sinxdx\int 6(\frac{x^2-4x-5}{x+1} + \sin{x})dx = 6\int \frac{x^2-4x-5}{x+1}dx + 6\int \sin{x}dx

=6(x5)(x+1)x+1dx+6(cosx)= 6\int \frac{(x-5)(x+1)}{x+1}dx + 6 (-\cos{x})
=6(x5)dx6cosx = 6\int (x-5)dx -6 \cos{x}
=6xdx65dx6cosx= 6 \int xdx - 6 \int 5dx- 6 \cos{x}
=6x2230x6cosx=6 \cdot \frac{x^2}{2} - 30x -6 \cos{x}
=3x230x6cosx+C=3x^2 - 30x -6 \cos{x} + C

Example 2

Evaluate lnxx4dx\int \frac{\ln{x}}{x^4}dx.

Solution: We’ll use integration by parts for this problem. Let’s choose u=lnxu = \ln{x} and dv=1x4dx=x4dxdv=\frac{1}{x^4}dx = x^{-4}dx. Differentiating uu to find dudu and integrating dvdv to find vv, we have:

du=1xdxdu = \frac{1}{x}dx
andand
v=x4dx=13x3v = \int x^{-4}dx = \frac{-1}{3x^3}

Now, using the integration by parts formula:

udv=uvvdu\int udv = uv - \int vdu
lnxx4dx=lnx13x313x31xdx\int \frac{\ln{x}}{x^4}dx = \ln{x} \cdot \frac{-1}{3x^3} - \int \frac{-1}{3x^3} \cdot \frac{1}{x}dx
=lnx3x3+13x4dx= \frac{-\ln{x}}{3x^3} + \frac{1}{3}\int x^{-4}dx
=lnx3x3+13x33=\frac{-\ln{x}}{3x^3} + \frac{1}{3} \cdot \frac{x^{-3}}{-3}
=lnx3x319x3+C=\frac{-\ln{x}}{3x^3} - \frac{1}{9x^3} + C

Example 3

Evaluate sin(lnx)xdx\int \frac{\sin{(\ln x)}}{x}dx.

Solution: We can solve this problem with u-substitution. Let u=lnxu = \ln x. Then du=1xdu = \frac{1}{x}.

sin(lnx)xdx=sinudu\int \frac{\sin{(\ln x)}}{x}dx = \int \sin{u}du
=cosu+C=-\cos{u}+C
=cos(lnx)+C=-\cos{(\ln x)} + C

Example 4

Evaluate x+13x2+10x+3dx\int \frac{x+1}{3x^2+10x+3}dx.

Solution: We can solve this problem with integration by partial fractions.

x+13x2+10x+3=x+1(x+3)(3x+1)\frac{x+1}{3x^2+10x+3} = \frac{x+1}{(x+3)(3x+1)}
=Ax+3+B3x+1= \frac{A}{x+3} + \frac{B}{3x+1}
=A(3x+1)+B(x+3)(x+3)(3x+1)=\frac{A(3x+1)+B(x+3)}{(x+3)(3x+1)}
=3Ax+A+Bx+3B(x+3)(3x+1)=\frac{3Ax+A+Bx+3B}{(x+3)(3x+1)}
=3Ax+Bx+A+3B(x+3)(3x+1)=\frac{3Ax+Bx+A+3B}{(x+3)(3x+1)}

Now, we set up our two equations. One represents the x terms, and one represents the constants.

So our equations are 1=3A+B1=3A+B and 1=A+3B1=A + 3B.

Solving this system of equations, we find that A=14A = \frac{1}{4} and B=14B=\frac{1}{4}.

Now we can evaluate our integral using the second Reciprocal Rule.

x+13x2+10x+3dx=(14(x+3)+14(3x+1))dx\int\frac{x+1}{3x^2+10x+3}dx=\int(\frac1{4(x+3)}+\frac1{4(3x+1)})dx
=141x+3dx+1413x+1dx=\frac{1}{4}\int \frac{1}{x+3}dx+\frac{1}{4}\int \frac{1}{3x+1}dx
=14ln(x+3)+112ln(3x+1)+C=\frac{1}{4} \ln{(x+3)}+\frac{1}{12} \ln{(3x+1)} + C

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