Definite Integrals

Definite integrals generalize the rule for finding the areas under the curves using integrals.  In the past, we’ve learned how to work on the antiderivative and indefinite integrals of continuous functions. Now, it’s time to explore how to establish the rules and properties when the limits are defined and boundaries are set for the input values.

In this article, we’ll show you the core properties and formulas needed to evaluate and apply definite integrals. By the end of our discussion, we want you to feel confident when working with definite integrals and their applications.

What Are Definite Integrals?

Definite integrals represent the areas found under the curve bounded between certain intervals. Mathematically, we define the definite integral of f(x) for the interval, x \in[a, b] as shown below.

\begin{aligned}\textbf{Definite Integral} = \boldsymbol{\int_{a}^{b} f(x) \phantom{x}dx}\end{aligned}

We’re back with the area under the curve, but this time, we have the interval, [a, b]. We can estimate the area of the curve by dividing it with rectangles that share a uniform width, h = \dfrac{b - a}{n}. Adding these sums will lead to the area of the curve – similar to how we use the Riemann sum to estimate indefinite integrals. 

This means that when asked to calculate the area under the curve of a function, y = f(x), bounded by the limits, x = a and x =b, we are simply evaluating the definite integral of f(x) between the interval of x \in [a, b]

Definite Integral Formula (Limit Sum)

We can define definite integrals in two ways: as an estimate of all the rectangles’ sums through Riemann’s sum or using the fundamental theorem of calculus relates definite and indefinite integrals. We’ve briefly discussed how we can extend Riemann’s sum to calculate the area of the curve and the function’s definite integral, so it’s time for us to establish its first formula.

\begin{aligned}\int_{a}^{b} f(x) \phantom{x}dx &= \lim_{n \rightarrow \infty} S_n\\&= \lim_{n \rightarrow \infty}[f(x_1)h + f(x_2)h +f(x_3)h + ...+ f(x_n)h]\\&= \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} f(x_i)h\end{aligned}

Of course, this mathematical definition is most helpful when evaluating the indefinite integral of the function is challenging. Numerical approximations of definite integrals highly depend on these approximations’ formulas to calculate complex definite integrals.

Definite Integral Formula (Fundamental Theorem of Calculus)

 Now let’s focus on a formula that’s easier for us to remember and work on. By lifting the limitations for our function to be continuous as well as positive, we can now establish the mathematical definition for the function’s definite integral. 

When the function, f(x), is bounded by the lower and upper limits: x = a and x =b, respectively.  

\begin{aligned}\int_{a}^{b} f(x) \phantom{x}dx &= F(b) - F(a)\\&= F(x)|_{a}^{b},\\&\text{where } F(x) = \int f(x) \phantom{x}dx\\&\phantom{xxxxx}F^{\prime}(x)= f(x) \end{aligned}

This formula can be established using the fundamental theorem of calculus. It simplifies our understanding of definite and gives us a better approach to evaluating definite integrals. For the rest of our discussion, we’ll focus our attention on the second formula.

How To Evaluate Definite Integral?

When evaluating definite integrals, we simply evaluate the indefinite integral of the function. Suppose that we’re working with \int_{a}^{b} f(x)\phantom{x} dx:

  1. We simply find the expression for F(x) = \int f(x) \phantom{x}dx
  2. Evaluate F(b) and F(a).
  3. Subtract these two expressions to find the definite integral of f(x) from x = a to x = b.

The best way to master the process is through practice. Let’s begin by trying out simple examples first.

Problem 1

Evaluate the definite integral, \int_{0}^{4} x^2 \phantom{x}dx.

Let’s first work on \int x^2 \phantom{x}dx by applying the power rule, \int x^n \phantom{x}dx = \dfrac{x^{n + 1}}{n + 1} + C.

\begin{aligned}\int x^2 \phantom{x}dx &= \dfrac{x^{2 +1}}{2 + 1} +C\\&= \dfrac{x^3}{3} +C\end{aligned}

Now that we have the antiderivative of our function, F(x) = \dfrac{x^3}{3} + C, we simply evaluate the function at x = 0 and x =4.

\boldsymbol{x = 0}\boldsymbol{x = 4}
\begin{aligned}F(0) &= \dfrac{0^3}{3}\\&= 0\end{aligned}\begin{aligned}F(4) &= \dfrac{4^3}{3}\\&= \dfrac{64}{3}\end{aligned}

To find the definite integral’s value, we simply subtract these two values.

\begin{aligned}\int_{0}^{4} x^2 \phantom{x}dx &= \dfrac{x^3}{3}|_{0}^{4}\\&= \dfrac{64}{3} - 0\\&= \dfrac{64}{3} \end{aligned}

Hence, we have \int_{0}^{4} x^2 \phantom{x}dx = \dfrac{64}{3}.

Now, curious about what happened with the constant, C? Before we’ve been constantly reminded to account for the arbitrary constant, C, but since we are dealing with definite integrals, these get “canceled out”.

\begin{aligned}\int_{a}^{b} f(x) \phantom{x}dx &= F(x) + C |_{a}^{b}\\&= [F(b) + C] - [F(a) + C]\\&= F(b) + \cancel{C }- F(a) - \cancel{C}\\&= F(b) -F(a)\end{aligned}

Of course, if we write definite integrals’ evaluations this way, it becomes tedious over time. This is why moving forward; we can disregard the arbitrary constant when dealing with definite integrals.

Definite Integral Properties and Formula

Before we work on more complex examples, we want to show you some of the important definite integral properties to remember. You may have encountered some of these when working with antiderivatives and indefinite integrals as well.

1) Sum of Difference Property: \boldsymbol{\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 }

This means that we can distribute the definite integral operation to break down complex functions.

2) Constant Multiple Property: \boldsymbol{\int_{a}^{b} [c\cdot f(x)]\phantom{x}dx = c\int_{a}^{b} f(x) \phantom{x}dx }

To easily manipulate definite integrals where the function’s terms has common factors, we can factor out the common factor then evaluate the resulting function instead.

3) Reverse Interval Property: \boldsymbol{\int_{a}^{b} f(x)\phantom{x}dx = -\int_{b}^{a} f(x) \phantom{x}dx}

When the limits are reversed, the resulting definite integral will be the negative of our original expression’s value.

4) Zero-Length Interval Property: \boldsymbol{\int_{a}^{b} f(x)\phantom{x}dx = -\int_{b}^{a} f(x) \phantom{x}dx}

When the lower and upper limits are identical, the definite integral’s value will always be equal to zero.

5) Combining Interval Property: \boldsymbol{\int_{a}^{b} f(x)\phantom{x}dx + \int_{b}^{c} f(x)\phantom{x}dx = \int_{a}^{c} f(x)\phantom{x}dx}

When given two definite integrals sharing the same lower limit and upper limit and function, we can simplify the expression into one definite integral by combining the intervals. 

These properties will come in handy when working with complex functions’ definite integrals. For now, let’s try applying some of these properties by trying out the problems below.

Problem 2

Suppose that \int_{-2}^{4} f(x) \phantom{x}dx = -4 and \int_{-2}^{4} g(x) \phantom{x}dx = 6, determine the values of the following:

a. \int_{-2}^{4} [5f(x) – 6g(x)] \phantom{x}dx

We can distribute definite integral within the expression and take out the respective factors.

\begin{aligned}\int_{-2}^{4} [5f(x) - 6g(x)] \phantom{x}dx &= \int_{-2}^{4} 5f(x) \phantom{x}dx - \int_{-2}^{4} 6g(x) \phantom{x}dx\\&= 5\int_{-2}^{4} f(x) \phantom{x}dx - 6\int_{-2}^{4} g(x) \phantom{x}dx\end{aligned}

Substitute the given values for these definite integrals to simplify the given expression.

\begin{aligned}\int_{-2}^{4} [5f(x) - 6g(x)] \phantom{x}dx &= 5(-4) - 6(6)\\&= -20 -36\\&= -56\end{aligned}

b. \int_{-2}^{4} f(x) - \int_{4}^{-2} 3g(x) \phantom{x}dx

For this problem, our second expression’s lower and upper limits’ positions reverse. Apply the reverse interval property to rewrite our expression.

\begin{aligned}\int_{-2}^{4} f(x) - \int_{4}^{-2} 3g(x) \phantom{x}dx &=\int_{-2}^{4} f(x) \phantom{x}dx - \left(-\int_{-2}^{4} 3g(x) \right ) \phantom{x}dx\\&=\int_{-2}^{4} f(x) \phantom{x}dx + \int_{-2}^{4} 3g(x)  \phantom{x}dx \end{aligned}

Now, factor out 3 in the second term and substitute appropriate values to simplify the definite integrals.

\begin{aligned}\int_{-2}^{4} f(x) – \int_{4}^{-2} 3g(x) \phantom{x}dx &=\int_{-2}^{4} f(x) \phantom{x}dx + 3\int_{-2}^{4} g(x)  \phantom{x}dx\\&= -4 + 3(6)\\&= 14 \end{aligned}

This means that \int_{-2}^{4} f(x) - \int_{4}^{-2} 3g(x) \phantom{x}dx is simply equal to 14.

Now that we’ve shown you the important definite integral properties, why don’t we try evaluating more complex definite integrals?

Problem 3

Evaluate the definite integral, \int_{-2}^{2} (x^3 – 4x + 3) \phantom{x}dx.

We can either apply the definite integral properties or work on the antiderivative first. For now, let us show you how the properties we’ve discussed can be applied in evaluating definite integrals. Apply the power and constant multiple formulas for integrals when necessary. 

Method 1:

\begin{aligned}\int_{-2}^{2} (x^3 - 4x + 3) \phantom{x}dx &= \int_{-2}^{2} x^3 \phantom{x}dx- \int_{-2}^{2}4x \phantom{x}dx+ \int_{-2}^{2}3 \phantom{x}dx\\&= \int_{-2}^{2} x^3 \phantom{x}dx- 4\int_{-2}^{2}x \phantom{x}dx+ \int_{-2}^{2}3 \phantom{x}dx\\&= \left[\dfrac{x^4}{4} \right ]_{-2}^{2} -4 \left[\dfrac{x^2}{2} \right ]_{-2}^{2} + [3x]_{-2}^{2}\\&= \left(\dfrac{2^4}{4} -\dfrac{(-2)^4}{4} \right ) -4(2^2 - (-2)^2) + 3(2 -(-2))\\&= 12\end{aligned}

Another way to evaluate the definite integral, \int_{-2}^{2} (x^3 – 4x + 3) \phantom{x}dx, is by working on its antiderivative first.  Apply appropriate integral formulas when needed.

Method 2:

\begin{aligned}\int (x^3 - 4x + 3) \phantom{x}dx &=\int x^3\phantom{x}dx-\int 4x\phantom{x}dx + \int 3\phantom{x}dx\\&= \int x^3\phantom{x}dx- 4\int x\phantom{x}dx + \int 3\phantom{x}dx\\&= \dfrac{x^4}{4}- 4\cdot \dfrac{x^2}{2}+ \int 3\phantom{x}dx\\&= \dfrac{1}{4}x^4 - 2x^2 + 3x +C  \end{aligned}

Once we have our function’s antiderivative, we’ll simply evaluate the F(x) at their lower and upper limits then subtract the results.

\begin{aligned}\int_{-2}^{2} (x^3 - 4x + 3) \phantom{x}dx &=\left[\dfrac{1}{4}x^4 - 2x^2 + 3x \right ]_{-2}^{2}\\&= \left[\dfrac{1}{4}(2)^4 - 2(2)^2 + 3(2) \right ]-\left[\dfrac{1}{4}(-2)^4 - 2(-2)^2 + 3(-2) \right ]\\&= 12\end{aligned}

This example shows that regardless of the process or method applied, the definite integral’s value will remain the same. For the next example though, we’ll only use one method. What you can do is try using the remaining method and compare your final answer!

Problem 4

Evaluate the definite integral, \int_{1}^{4} \dfrac{x}{(x^2 + 4)^3} \phantom{x}dx.

By inspecting the integrand alone, we can see that it’ll be much simpler if we work on the antiderivative first. Integrating our expression will require a substitution method, but don’t worry, we’ll break down the steps for you here.

\begin{aligned}u &= x^2 + 4\\du &= 2x \phantom{x}dx\\\dfrac{1}{2}\phantom{x}du &= x \phantom{x}dx\end{aligned}

We can rewrite the integrand in terms of u now. Keep in mind that the limits will also change form x = 1\Rightarrow u =5 and x = 4\Rightarrow u =20.

\begin{aligned}\int_{1}^{4} \dfrac{x}{(x^2 + 4)^3} \phantom{x}dx&= \int_{1}^{4} \dfrac{1}{(x^2 + 4)^3} \phantom{x} xdx\\&=\int_{5}^{20} \dfrac{1}{2} \dfrac{1}{u^3} \phantom{x}du\\&=\dfrac{1}{2} \int_{5}^{8} \dfrac{1}{u^3} \phantom{x}du\\&=\dfrac{1}{2} \int_{5}^{20} u^{-3} \phantom{x}du\end{aligned}

Now, simplify the antiderivative first then evaluate the expression at the new limits: u = 5 and u = 8.

\begin{aligned}\dfrac{1}{2} \int u^{-3} \phantom{x}du &=  \dfrac{1}{2} \cdot \dfrac{u^{-3 + 1}}{-3 + 1} \\&= -\dfrac{u^{-2}}{4}\\\\\dfrac{1}{2} \int_{5}^{20} u^{-3}&= \left[-\dfrac{u^{-2}}{4} \right ]_{5}^{20}\\&= -\dfrac{1}{4}[(20)^{-2} - (5)^{-2}]\\&= \dfrac{3}{320}\end{aligned}

Hence, we have \int_{1}^{4} \dfrac{x}{(x^2 + 4)^3} \phantom{x}dx = \dfrac{3}{320}.