Integration might seem tricky. But it’s just the opposite of differentiation. In this article, we’ll review the basics of integration and solve some basic integration problems requiring only high school algebra.
An understanding of differentiation is necessary to solve integrals, so if you haven’t already, please check out my guide on how to differentiate a function step by step.
What is Integration?
Integrals calculate areas, accumulated values, and other quantities in physics and engineering.
In math, integration is denoted
$$\int f(x) dx = F(x) + C.$$
Where ∫ is the integral symbol. f(x) is the integrand, dx is the differential, F(x) is the antiderivative, and C is the constant of integration.
Indefinite vs. Definite Integrals
Indefinite integrals give a general function, which, when differentiated, gives the integrand. Definite integrals calculate the area between two points called the limits of integration. For now, we’ll focus on indefinite integrals since they’re also the foundation for computing definite integrals.
Basic Integration Rules
Before solving problems, let’s review some key rules for integration, similar to the basic differentiation rules.
Constant Rule
$$\int c \, dx = cx + C$$
If you’re integrating a constant, multiply it by x and add a constant of integration.
Power Rule
$$\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad \text{for } n \neq -1$$
Increase the exponent by 1, then divide by the new exponent. Don’t forget the constant of integration.
Linearity
Linearity combines the sum and difference rule and the constant multiple rule.
$$\int [cf(x) \pm kg(x)] dx = c\int f(x) dx \pm k\int g(x) dx$$
Now, let’s give an example of how to apply these rules.
Example 1: Compute the integrals:
(a) \(\int x^2 + 7x + 12 dx\)
Solution: We start by breaking up the integral using linearity:
$$\int x^2 \, dx + \int 7x \, dx + \int 12 \, dx.$$
Using the power rule gives,
$$\frac{x^3}{3} + \frac{7x^2}{2} + 12x + C.$$
(b) \(\int \frac{(x+1)^2}{x} dx\)
Solution: We start by expanding the numerator to obtain
$$\int \frac{x^2 + 2x + 1}{x} \, dx.$$
Next, split the fraction into three terms,
$$\int x + 2 + \frac{1}{x} dx.$$
Rewriting using linearity gives
$$\int x \, dx + \int 2 \, dx + \int \frac{1}{x} dx.$$
Finally, we integrate to arrive at
$$\frac{x^2}{2} + 2x + \ln|x| + C.$$
(c) \(\int \frac{x+2}{\sqrt{x}} dx\)
Solution: Breaking up the fraction and rewriting the roots in exponential form gives us
$$\int \frac{x}{\sqrt{x}} + \frac{2}{\sqrt{x}} dx$$
$$\int \frac{x}{x^{\frac{1}{2}} } + \frac{2}{x^{\frac{1}{2}}} dx$$
$$\int x^{\frac{1}{2}} + 2x^{-\frac{1}{2}} dx.$$
Using the power rule to integrate each term, we arrive at
$$\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 2 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C$$
$$\frac{2}{3} x^{\frac{3}{2}} + 4x^{\frac{1}{2}} + C.$$
Or, in radical form,
$$\frac{2}{3} \sqrt{x^{3}} + 4\sqrt{x} + C.$$
(d) \(\int \frac{x^2 + 7x + 12}{x^2 + 3x} dx\)
Solution: Factoring the numerator and the denominator gives,
$$\int \frac{(x+3)(x+4)}{x(x+3)} dx = \int \frac{x+4}{x} dx.$$
Split the fraction into two terms and use linearity to get
$$\int 1 + \frac{4}{x} dx$$
$$\int 1 \, dx + \int \frac{4}{x} dx.$$
Lastly, we integrate to arrive at
$$x + 4\ln|x| + C.$$
Integrating Common Functions
Now that we’ve covered the basics of integration, let’s explore how to integrate some standard functions you’ll encounter.
Polynomial Functions
This section is only here for completeness. All polynomials can be integrated using the techniques covered in the previous section. See part a of Example 1 for an example of computing the integral of a polynomial.
Trigonometric Functions
$$\int \sin(x) \, dx = -\cos(x) + C$$
$$\int \cos(x) \, dx = \sin(x) + C$$
$$\int \sec^2(x) \, dx = \tan(x) + C$$
$$\int \csc^2(x) \, dx = -\cot(x) + C$$
$$\int \sec(x)\tan(x) \, dx = \sec(x) + C$$
$$\int \csc(x)\cot(x) \, dx = -\csc(x) + C$$
Be careful with the signs.
Exponential Functions
$$\int e^x \, dx = e^x + C$$
Hyperbolic Functions
$$\int \sinh(x) \, dx = \cosh(x) + C$$
$$\int \cosh(x) \, dx = \sinh(x) + C$$
$$\int \text{sech}^2(x) \, dx = \tanh(x) + C$$
$$\int \text{csch}^2(x) \, dx = -\coth(x) + C$$
$$\int \text{sech}(x) \tanh(x) \, dx = -\text{sech}(x) + C$$
$$\int \text{csch}(x) \coth(x) \, dx = -\text{csch}(x) + C$$
Be careful with the signs
Other Elementary Functions
$$\int \frac{1}{x} \, dx = \ln|x| + C$$
$$\int \frac{1}{\sqrt{1 – x^2}} \, dx = \arcsin(x) + C$$
$$\int \frac{1}{1 + x^2} \, dx = \arctan(x) + C$$
$$\int \frac{1}{|x|\sqrt{x^2 – 1}} \, dx = \text{arcsec}(x) + C$$
Notice that we have not provided integration formulas for some trigonometric functions, inverse trigonometric functions, and logarithmic functions. This is because these functions require advanced techniques.
The substitution rule is needed to integrate tangent, secant, cotangent, and cosecant. Integration by parts is needed to integrate logarithmic and inverse trigonometric functions.
The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects the two main concepts in calculus: differentiation and integration, and consists of two parts.
The First Part of the Fundamental Theorem
The first part of the Fundamental Theorem tells us that if you take the derivative of the integral of a function, you get the original function back, or mathematically
$$ \frac{d}{dx}\int_a^x f(t) dt = f(x).$$
The Second Part of the Fundamental Theorem
If you have a function f(x) that’s continuous over an interval [a, b], the integral of f(x) from a to b is
$$\int_a^b f(x) dx = F(b) – F(a).$$
Where F(x) is the antiderivative (or integral) of f(x), and a and b are the limits of integration.
This tells us that we can calculate definite integrals by first finding the antiderivative and subtracting the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit.
Let’s give an example of how to compute definite integrals.
Example 2: Find the value of the definite integrals
(a) \(\int_0^{\frac{\pi}{2}} \cos{x} dx\)
(b) \(\int_0^1 e^x dx\)
Solution: For both problems, we only need to compute the antiderivative and apply part 2 of the Fundamental Theorem of Calculus.
(a) $$\int_0^{\frac{\pi}{2}} \cos{x} \, dx = \left[ \sin{x} \right]_0^{\frac{\pi}{2}} = \sin{\frac{\pi}{2}} – \sin{0} = 1 – 0 = 1.$$
(b) $$ \int_0^1 e^x \, dx = \left[ e^x \right]_0^1 = e^1 – e^0 =e – 1.$$
Conclusion
With the methods we’ve discussed, you can solve basic integration problems, which are the foundation for more advanced techniques like substitution and integration by parts. These methods can also be applied to solve more interesting integrals.
