Basic Integration Problems for Beginners
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.
Understanding 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.
The following infographic illustrates the concepts covered in this article.
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
Example 1: Compute the integrals:
(a) \(\int x^2 + 7x + 12 dx\)
(b) \(\int \frac{(x+1)^2}{x} dx\)
(c) \(\int \frac{x+2}{\sqrt{x}} dx\)
(d) \(\int \frac{x^2 + 7x + 12}{x^2 + 3x} dx\)
Solution: (a) \(\int x^2 + 7x + 12 dx\)
We start by breaking up the integral using linearity:
$$\int x^2 \, dx + 7 \int x \, dx + \int 12 \, dx.$$
Using the power rule gives a final answer of
$$\frac{1}{3}x^3 + \frac{7}{2}x^2 + 12x + C.$$
(b) \(\int \frac{(x+1)^2}{x} dx\)
We start by expanding the numerator to obtain
$$\int \frac{x^2 + 2x + 1}{x} \, dx.$$
Next, split the fraction into three terms to get
$$\int \frac{x^2}{x} + \frac{2x}{x} + \frac{1}{x} dx.$$
Simplifying, this is equivalent to
$$\int x + 2 + \frac{1}{x} dx.$$
Rewriting using linearity gives
$$\int x \, dx + \int 2 \, dx + \int \frac{1}{x} dx.$$
Finally, we integrate to arrive at a final answer of
$$\frac{1}{2}x^2 + 2x + \ln|x| + C.$$
(c) \(\int \frac{x+2}{\sqrt{x}} dx\)
Breaking up the fraction gives us.
$$\int \frac{x}{\sqrt{x}} + \frac{2}{\sqrt{x}} dx$$
Rewriting the roots in exponential form, we obtain
$$\int \frac{x}{x^{\frac{1}{2}} } + \frac{2}{x^{\frac{1}{2}}} dx$$
Rewriting the second term using negative exponents, we get
$$\int x^{\frac{1}{2}} + 2x^{-\frac{1}{2}} dx.$$
Using linearity, this is equivalent to
$$\int x^{\frac{1}{2}} dx + 2\int x^{-\frac{1}{2}} dx.$$
Using the power rule to integrate each term, we arrive at
$$\frac{2}{3} x^{\frac{3}{2}} + 4x^{\frac{1}{2}} + C.$$
Rewriting in radical form gives a final answer of
$$\frac{2}{3} \sqrt{x^{3}} + 4\sqrt{x} + C.$$
(d) \(\int \frac{x^2 + 7x + 12}{x^2 + 3x} dx\)
Factoring the numerator and the denominator gives,
$$\int \frac{(x+3)(x+4)}{x(x+3)} dx.$$
This is equivalent to
$$\int \frac{x+4}{x} dx.$$
Now, split the fraction into two terms to get
$$\int \frac{x}{x} + \frac{4}{x} dx.$$
Simplifying, the first term we obtain
$$\int 1 + \frac{4}{x} dx$$
Using linearity gives
$$\int 1 \, dx + 4\int \frac{1}{x} dx.$$
Lastly, we integrate to arrive at a final answer of
$$x + 4\ln|x| + C.$$
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.
Definite Integrals and 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.
Properties of Definite Integrals
Definite integrals have the following properties that are useful on occasion:
$$\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx, a \leq b \leq c$$
$$\int_{-a}^a f(x)dx = 2\int_0^a f(x)dx, f(-x) = f(x)$$
$$\int_{-a}^a f(x)dx = 0, f(-x) = -f(x).$$
Definite Integral Example
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: (a) $$\int_0^{\frac{\pi}{2}} \cos{x} dx.$$
Integrating gives
$$\sin{x} ]_0^{\frac{\pi}{2}}.$$
Applying the Fundamental Theorem of Calculus, we obtain
$$\sin{\frac{\pi}{2}} – \sin{0}.$$
This is equivalent to
$$1 – 0.$$
Subtracting, we arrive at a final answer of
$$1.$$
(b) $$ \int_0^1 e^x dx.$$
Integrating gives
$$ e^x ]_0^1.$$
Applying the Fundamental Theorem of Calculus, we obtain
$$ e^1 – e^0.$$
Evaluating this expression gives a final answer of
$$ e – 1.$$
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.
Further Reading
The Ultimate Step-by-Step Guide to Solving Integrals Using Substitution – Now that you can solve basic integrals, the next step is learning more advanced integration techniques. Substitution is the simplest of these and is necessary for learning further integration techniques.
Integration by Parts Explained with Examples: A Step-by-Step Guide – Integration by parts, while more difficult than substitution, can be learned knowing only the methods discussed in this article. However, some more advanced problems require substitution.
