Lesson Explainer: The Fundamental Theorem of Calculus: Evaluating Definite Integrals Mathematics • Higher Education

In this explainer, we will learn how to use the fundamental theorem of calculus to evaluate definite integrals.

Definite integrals are all about the accumulation or sum of a particular quantity and are closely related to antiderivatives. They provide us with a powerful tool to help us understand and model real-world phenomena, appearing in many disciplines from pure mathematics, with geometric applications such as surface area and volume, to physics when determining the mass of an object, the work done, or the pressure exerted on an object, to name just a few.

The definite integral of the function 𝑓(π‘₯) from π‘₯=π‘Ž to π‘₯=𝑏 can be interpreted as the signed area under the curve of 𝑓(π‘₯) from π‘₯=π‘Ž to π‘₯=𝑏; a visual representation of this integral is given in the following diagram.

So, how are definite integrals defined? Before we give the precise definition, we note that we can estimate the area under the curve for some function 𝑦=𝑓(π‘₯), bounded by π‘₯=π‘Ž and π‘₯=𝑏, by first splitting up the interval [π‘Ž,𝑏] into 𝑛 subintervals of equal width, [π‘₯,π‘₯]οƒοŠ±οŠ§οƒ for 𝑖=1,…,𝑛, as shown in the diagram.

This gives 𝑛 rectangles of equal width, Ξ”π‘₯, where the height of each rectangle is given by the value of the function at each point, 𝑓(π‘₯), from the right endpoint of each subinterval. The area of each rectangle is the product of this height and width, 𝑓(π‘₯)Ξ”π‘₯. We can estimate the area under the curve of 𝑓(π‘₯) by summing the areas of each rectangle as Areaβ‰ˆπ‘“(π‘₯)Ξ”π‘₯+β‹―+𝑓(π‘₯)Ξ”π‘₯=ο„šπ‘“(π‘₯)Ξ”π‘₯.οŠ§οŠοŠοƒοŠ²οŠ§οƒ

This is also known as the right Riemann sum. As the number of rectangles 𝑛 gets larger and the width Ξ”π‘₯ gets smaller, this estimate will get closer to the true area under the curve. In fact, the definite integral, which gives the exact area under the curve, is defined by taking the limit of this sum as the number of rectangles approaches infinity.

Definition: The Definite Integral

Given a function 𝑓 that is continuous and defined on the interval [π‘Ž,𝑏], we can divide the interval into 𝑛 subintervals [π‘₯,π‘₯]οƒοŠ±οŠ§οƒ of equal width, Ξ”π‘₯, and choose sample points π‘₯∈[π‘₯,π‘₯]βˆ—οƒοƒοŠ±οŠ§οƒ. The definite integral from π‘₯=π‘Ž to π‘₯=𝑏 is defined in terms of the Riemann sum as 𝑓(π‘₯)π‘₯=ο„šπ‘“ο€Ήπ‘₯Δπ‘₯,οŒ»οŒΊοŠβ†’βˆžοŠοƒοŠ²οŠ§βˆ—οƒdlim where π‘₯=π‘Ž+𝑖Δπ‘₯,Ξ”π‘₯=π‘βˆ’π‘Žπ‘›=π‘₯βˆ’π‘₯,οƒοƒοƒοŠ±οŠ§ provided that the limit exists and gives the same value for all sample points π‘₯∈[π‘₯,π‘₯]βˆ—οƒοƒοŠ±οŠ§οƒ.

It does not matter which sample point π‘₯βˆ—οƒ in the subinterval [π‘₯,π‘₯]οƒοŠ±οŠ§οƒ is taken to be. Since the difference or width of the summands Ξ”π‘₯β†’0, so does the difference between any two points in the interval. This is because the choice of π‘₯βˆ—οƒ is arbitrary, which may produce different Riemann sums, which converge to the same value. In particular, the common choices are given by the following:

  • If π‘₯=π‘₯βˆ—οƒοƒ, that is, the function 𝑓 is evaluated at the right endpoint of each subinterval, then we have the right Riemann sum. The definite integral in terms of this sum is 𝑓(π‘₯)π‘₯=ο„šπ‘“(π‘₯)Ξ”π‘₯.οŒ»οŒΊοŠβ†’βˆžοŠοƒοŠ²οŠ§οƒdlim This is the choice that most people use when finding a specific Riemann sum or definite integral, for simplicity, and it corresponds to the example above with an estimate of the area under the curve using 𝑛 equal-width rectangles and the diagram, with the limit as π‘›β†’βˆž.
  • If π‘₯=π‘₯βˆ—οƒοƒοŠ±οŠ§, that is, the function 𝑓 is evaluated at the left endpoint of each subinterval, then we have the left Riemann sum.
  • If π‘₯=(π‘₯+π‘₯)2βˆ—οƒοƒοƒοŠ±οŠ§, that is, the function 𝑓 is evaluated at the midpoint of each subinterval, then we have the midpoint Riemann sum.

The definite integral always gives the signed area under the curve; the area given by the definite integral above the π‘₯-axis is always positive, while below the π‘₯-axis it is always negative, as shown in the diagram.

If there are parts of the curve that are below and above the π‘₯-axis in the interval [π‘Ž,𝑏], then the definite integral will be the area above the π‘₯-axis minus the area below the π‘₯-axis, within the interval [π‘Ž,𝑏].

So, how do we evaluate these definite integrals? Using the definition of the definite integral given in terms of the limit of Riemann sums would be cumbersome in practice. We can instead evaluate them by using the fundamental theorem of calculus.

The first part of the theorem allows us to determine the antiderivative from its indefinite integral, when a real-valued function is continuous on an interval [π‘Ž,𝑏]. Let’s first recall the first part of the fundamental theorem of calculus, concerning the existence of an antiderivative.

The First Part of the Fundamental Theorem of Calculus

If 𝑓 is a continuous real-valued function defined on [π‘Ž,𝑏] and we let 𝐹 be the function defined, for all π‘₯ in [π‘Ž,𝑏], as 𝐹(π‘₯)=𝑓(𝑑)𝑑,ο—οŒΊd then 𝐹 is uniformly continuous on [π‘Ž,𝑏] and differentiable on ]π‘Ž,𝑏[, and 𝐹′(π‘₯)=𝑓(π‘₯), for all π‘₯ in ]π‘Ž,𝑏[.

In other words, we can compute the antiderivative 𝐹 of some function 𝑓 by computing the indefinite integral of 𝑓 given as 𝐹(π‘₯)=𝑓(π‘₯)π‘₯+𝐢,d where 𝐢 is known as the constant of integration. By the first part of the theorem, antiderivatives of 𝑓 always exist when 𝑓 is continuous and there are infinitely many antiderivatives for 𝑓, obtained by adding this arbitrary constant to 𝐹.

We note that the constant of integration 𝐢 is included in the first term, which we usually add after integrating 𝑓(π‘₯), but we have explicitly stated it here to examine the indefinite integral. This is to elucidate the fact that there can be infinitely many antiderivatives parametrized by this constant. However, for definite integrals, we can ignore this constant or set it equal to zero since it is cancelled out, as we shall see.

The first part of the fundamental theorem of calculus also provides a powerful corollary, which we will use to evaluate definite integrals.

Corollary

The fundamental theorem is often employed to compute the definite integral of a function 𝑓 for which an antiderivative 𝐹 is known. Specifically, if 𝑓 is a real-valued continuous function on [π‘Ž,𝑏] and 𝐹 is an antiderivative of 𝑓 on [π‘Ž,𝑏], then 𝑓(𝑑)𝑑=[𝐹(𝑑)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

The square brackets [𝐹(𝑑)] are often used as a shorthand, after integration, to indicate the bounds the antiderivative is to be evaluated at and are equivalent to 𝐹(𝑏)βˆ’πΉ(π‘Ž).

The corollary assumes continuity on the whole interval as it follows from the first part of the fundamental theorem of calculus. This result is strengthened slightly in the second part of the theorem.

The Second Part of the Fundamental Theorem of Calculus (Newton–Leibniz Axiom)

Let 𝑓 be a real-valued function on a closed interval [π‘Ž,𝑏] and 𝐹 be an antiderivative of 𝑓 on [π‘Ž,𝑏]: 𝐹′(π‘₯)=𝑓(π‘₯).

If 𝑓 is Riemann integrable on [π‘Ž,𝑏], then 𝑓(𝑑)𝑑=[𝐹(𝑑)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

The second part of the fundamental theorem of calculus is somewhat stronger than the corollary because it does not assume that 𝑓 is continuous.

Though not strictly required by the second part, we will assume that all functions are continuous on [π‘Ž,𝑏] for the purpose of this explainer so that we can always determine the antiderivative for the integral to be valid. In fact, the corollary of the first part of the fundamental theorem of calculus is sufficient to evaluate definite integrals for our purposes, by this assumption of continuity.

Let’s consider a real-world application to give us a bit of intuition. Suppose that the temperature 𝑇(𝑑) of a cup of coffee is decreasing at a rate of π‘Ÿ(𝑑) degrees Celsius per minute, where π‘Ÿ(𝑑) is a continuous function.

In other words, the rate of change of the temperature is given by the first derivative of the temperature at time 𝑑: π‘Ÿ(𝑑)=𝑇𝑑.dd

At time 𝑑=0 (the start), the temperature of the cup of coffee is 40 degrees Celsius. How do we find the amount by which the temperature has decreased 5 minutes after the start? We can solve this problem using definite integration with the fundamental theorem of calculus, ο„Έπ‘Ÿ(𝑑)𝑑=𝑇(5)βˆ’π‘‡(0),d or find a general expression for the temperature 𝑇(𝑑) as 𝑇(𝑑)=ο„Έπ‘Ÿ(𝑑)𝑑+𝑇(0).d

In fact, for any quantity πœ†(𝑑) whose rate is given by the continuous function π‘Ÿ(𝑑), the definite integral, ο„Έπ‘Ÿ(𝑑)𝑑=πœ†(𝑏)βˆ’πœ†(π‘Ž),d describes the amount by which the quantity πœ† changed between 𝑑=π‘Ž and 𝑑=𝑏.

Definite integrals also satisfy certain properties, similar to indefinite integrals, derivatives, and limits. Let’s introduce some properties that will be useful for the problems in this explainer.

Properties of Definite Integrals

For functions 𝑓 and 𝑔 that are continuous on [π‘Ž,𝑏], we have the following:

  • The variable π‘₯ that appears in definite integrals is called the dummy variable, and we can replace this with another to get the same result: 𝑓(π‘₯)π‘₯=𝑓(𝑑)𝑑.dd
  • The definite integral of a constant π‘βˆˆβ„ is proportional the width of the interval: 𝑐π‘₯=𝑐(π‘βˆ’π‘Ž).d
  • We can split up definite integrals with a sum or difference: ο„Έ(𝑓(π‘₯)±𝑔(π‘₯))π‘₯=𝑓(π‘₯)π‘₯±𝑔(π‘₯)π‘₯.ddd
  • We can factor out a constant π‘βˆˆβ„ from definite integrals: 𝑐𝑓(π‘₯)π‘₯=𝑐𝑓(π‘₯)π‘₯.dd
  • We can also split up the integral with the limits for some value π‘βˆˆ[π‘Ž,𝑏] as 𝑓(π‘₯)π‘₯=𝑓(π‘₯)π‘₯+𝑓(π‘₯)π‘₯.ddd

These and other properties of definite integrals will be explored further in more detail in another explainer; we only state those that are useful for evaluating definite integrals for the problems in this explainer. They can be shown directly from the fundamental theorem of calculus; for instance, on the last property, we can split up the integral and apply the fundamental theorem of calculus to obtain 𝑓(π‘₯)π‘₯=𝑓(π‘₯)π‘₯+𝑓(π‘₯)π‘₯=(𝐹(𝑐)βˆ’πΉ(π‘Ž))+(𝐹(𝑏)βˆ’πΉ(𝑐))=𝐹(𝑏)βˆ’πΉ(π‘Ž).ddd

This is an intuitive property, since the areas of each of the parts add up to the total area over [π‘Ž,𝑏]. This can be visualized as follows:

We also note that when we find the antiderivative for definite integrals, we can ignore the constant of integration or set it to zero, since this is cancelled when finding the difference evaluated at the limits of integration, 𝐹(𝑏)βˆ’πΉ(π‘Ž).

In order to see this in action, consider the definite integral of 𝑓(π‘₯)=π‘₯ from π‘₯=0 to π‘₯=6, as shown on the plot of the function between these values. We will compute the area under the curve by using the fundamental theorem of calculus, but also graphically for this special case.

Since the curve, within the interval [0,6], lies above the π‘₯-axis, we expect the definite integral to be positive since this gives the signed area under the curve.

As 𝑓(π‘₯) is a linear function, it is continuous for all points π‘₯∈[0,6]; since it is continuous everywhere on ℝ, we can use the corollary of the first part of the fundamental theorem of calculus or the second part to evaluate this definite integral by using the antiderivative 𝐹(π‘₯) defined by 𝐹′(π‘₯)=𝑓(π‘₯), evaluating at the limits of integration π‘₯=0 and π‘₯=6, and finding the difference. The antiderivative is given by the indefinite integral of 𝑓(π‘₯) which we can find by using the power rule for integration as 𝐹(π‘₯)=ο„Έπ‘₯π‘₯=12π‘₯+𝐢.d

Therefore, the definite integral is given by ο„Έπ‘₯π‘₯=12π‘₯+𝐢=ο€Ό12(6)+πΆοˆβˆ’ο€Ό12(0)+𝐢=362=18.d

Hence, the area under the curve 𝑦=π‘₯ between 6 and 0 is 18 area units. As we can see, since the same constant of integration appears in both parts in the difference from the antiderivative, it is always cancelled, and thus we can ignore the constant of integration or set it to zero for definite integrals.

For this particular function, we can also compute the area under the curve graphically since it is the same as the area of the right triangle. Recall that a right triangle with base 𝑏 and height β„Ž has an area of 𝐴=12π‘β„Ž.

The right triangle, as shown in the plot, has a base and height of 6, and thus the area is 𝐴=12Γ—6Γ—6=18.

This gives the same result from the fundamental theorem of calculus, as expected.

If we had a constant function 𝑓(π‘₯)=𝑐, then the definite integral would be the signed area under the curve between π‘₯=π‘Ž and π‘₯=𝑏, equal to the area of the rectangle with lengths |𝑐| and |π‘βˆ’π‘Ž|, up to a sign.

This only works for constant and linear functions, since the area under the curve is equivalent to the area of the rectangle or triangle within [π‘Ž,𝑏]. For other polynomials or functions, we have to compute the area by using the definite integral with the fundamental theorem of calculus.

Now, let’s consider the definite integral of 𝑓(π‘₯)=π‘₯+2π‘₯ from π‘₯=1 to π‘₯=2, as shown on the plot of the function between these values.

Since the curve, within the interval [1,2], lies above the π‘₯-axis, we expect the definite integral to be positive since this gives the signed area under the curve.

As 𝑓(π‘₯) is a polynomial, it is continuous for all points π‘₯∈[1,2], since it is continuous everywhere on ℝ. As before, by using the fundamental theorem of calculus, we can compute the antiderivative: 𝐹(π‘₯)=ο„Έο€Ήπ‘₯+2π‘₯π‘₯=13π‘₯+π‘₯+𝐢.d

Therefore, the definite integral is given by ο„Έο€Ήπ‘₯+2π‘₯π‘₯=13π‘₯+π‘₯+𝐢=ο€Ό13(2)+(2)+πΆοˆβˆ’ο€Ό13(1)+(1)+𝐢=ο€Ό83+4+πΆοˆβˆ’ο€Ό13+1+𝐢=163.d

Hence, the area under the curve 𝑦=π‘₯+2π‘₯ between 1 and 2 is 163 area units.

Now, consider the definite integral of 𝑓(π‘₯)=(3π‘₯)sin from π‘₯=0 to π‘₯=πœ‹, as shown in the plot.

Since part of the curve, within the interval [0,πœ‹], lies above the π‘₯-axis and another part lies below the π‘₯-axis, the definite integral will give the area above the π‘₯-axis minus the area under the π‘₯-axis, which we expect to be positive.

The function 𝑓(π‘₯) is continuous for all points π‘₯∈[0,πœ‹], since it is continuous everywhere on ℝ. Again, by using the fundamental theorem of calculus, we can compute the antiderivative: 𝐹(π‘₯)=ο„Έ(3π‘₯)π‘₯=βˆ’13(3π‘₯)+𝐢;sindcos then, we evaluate it at the limits of integration and find the difference: ο„Έ(3π‘₯)π‘₯=ο”βˆ’13(3π‘₯)=ο€Όβˆ’13(3(πœ‹))οˆβˆ’ο€Όβˆ’13(3(0))=ο€Όβˆ’13(βˆ’1)+ο€Ό13(1)=23.οŽ„οŠ¦οŽ„οŠ¦sindcoscoscos

Since the definite integral gives the signed area under the curve, this is the area in red subtracted from the area in blue. In other words, it is the area under the curve above the π‘₯-axis minus the area under the curve below the π‘₯-axis, in the interval [0,πœ‹].

Consider the definite integral of 𝑓(π‘₯)=1+𝑒οŠͺ from π‘₯=0 to π‘₯=2ln, as shown in the plot below.

Since the curve within the interval [0,2]ln lies above the π‘₯-axis, we expect the definite integral to be positive since this gives the signed area under the curve.

The function 𝑓(π‘₯) is continuous for all points π‘₯∈[0,2]ln, since it is continuous everywhere on ℝ. Using the fundamental theorem of calculus, we can evaluate this definite integral from its antiderivative: ο„Έο€Ή1+𝑒π‘₯=π‘₯+14𝑒=ο€Ό(2)+14π‘’οˆβˆ’ο€Ό(0)+14π‘’οˆ=(2+4)βˆ’ο€Ό14=154+2.lnlnlnοŠͺοŠͺο—οŠ¨οŠ¦οŠͺ()οŠͺ()dlnlnln

Now, let’s look at a few examples to practice and help strengthen our understanding. In the first example, we will evaluate the definite integral of a quadratic function.

Example 1: Evaluating the Definite Integral of a Quadratic Function

Let 𝑓(π‘₯)=6π‘₯+1. Evaluate the definite integral of 𝑓 from π‘₯=2 to π‘₯=3.

Answer

In this example, we want to find the definite integral of 𝑓(π‘₯)=6π‘₯+1 from π‘₯=2 to π‘₯=3. Since the curve, within the interval [2,3], lies above the π‘₯-axis, we expect the definite integral to be positive since this gives the signed area under the curve. This is visually represented in the plot, which shows the area under the curve between π‘₯=2 and π‘₯=3.

We begin by first finding the antiderivative of 𝑓(π‘₯)=6π‘₯+1 from the indefinite integral by using the power rule for integration: 𝐹(π‘₯)=ο„Έο€Ή6π‘₯+1π‘₯=2π‘₯+π‘₯+𝐢.d

For the definite integral, we can apply the fundamental theorem of calculus, which states that if 𝑓 is continuous on [π‘Ž,𝑏] and 𝐹′(π‘₯)=𝑓(π‘₯), then 𝑓(π‘₯)π‘₯=[𝐹(π‘₯)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

We note that we can ignore the constant of integration for the antiderivative 𝐹(π‘₯), since this is cancelled in the difference 𝐹(𝑏)βˆ’πΉ(π‘Ž).

The integrand, for our integral, is a polynomial and hence continuous and defined for all points π‘₯∈[2,3]. Therefore, evaluating the antiderivative at the limits of integration and finding their difference, we have ο„Έο€Ή6π‘₯+1π‘₯=2π‘₯+π‘₯=ο€Ί2(3)+3ο†βˆ’ο€Ί2(2)+2=57βˆ’18=39.d

In our next example, we will evaluate the definite integral of a function involving exponential and trigonometric functions.

Example 2: Evaluating the Definite Integral of a Function Involving Exponential and Trigonometric Functions

Evaluate ο„Έ(2π‘₯βˆ’3𝑒)π‘₯οŠ¨οŠ¦ο—sind.

Answer

In this example, we have to evaluate the definite integral of 2π‘₯βˆ’3𝑒sin from π‘₯=0 to π‘₯=2. Since the curve, within the interval [0,2], lies below the π‘₯-axis, we expect the definite integral to be negative since this gives the signed area under the curve. This is visually represented in the plot, which shows the area under the curve between π‘₯=0 and π‘₯=2.

We can find the indefinite integral by first finding the antiderivative of the integrand 2π‘₯βˆ’3𝑒sin, which gives ο„Έ(2π‘₯βˆ’3𝑒)π‘₯=βˆ’2π‘₯βˆ’3𝑒+𝐢.sindcos

For the definite integral, we can apply the fundamental theorem of calculus, which states that if 𝑓 is continuous on [π‘Ž,𝑏] and 𝐹′(π‘₯)=𝑓(π‘₯), then 𝑓(π‘₯)π‘₯=[𝐹(π‘₯)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

We note that we can ignore the constant of integration for the antiderivative 𝐹(π‘₯), since this is cancelled in the difference 𝐹(𝑏)βˆ’πΉ(π‘Ž).

The integrand, for our integral, is the difference of continuous functions and hence is continuous and defined for all points π‘₯∈[0,2]. Therefore, evaluating the antiderivative at the limit of integration and finding their difference, we have ο„Έ(2π‘₯βˆ’3𝑒)π‘₯=[βˆ’2π‘₯βˆ’3𝑒]=ο€Ήβˆ’2(2)βˆ’3π‘’ο…βˆ’ο€Ήβˆ’2(0)βˆ’3𝑒=βˆ’3π‘’βˆ’22+5.οŠ¨οŠ¦ο—ο—οŠ¨οŠ¦οŠ¨οŠ¦οŠ¨sindcoscoscoscos

In our next example, we will evaluate the definite integral of a function involving a root.

Example 3: Evaluating the Definite Integral of a Root Function

Evaluate ο„Έβˆ’2√π‘₯π‘₯οŠͺd.

Answer

In this example, we have to evaluate the definite integral of βˆ’2√π‘₯ from π‘₯=4 to π‘₯=9. Since the curve, within the interval [4,9], lies below the π‘₯-axis, we expect the definite integral to be negative since this gives the signed area under the curve. This is visually represented in the plot, which shows the area under the curve between π‘₯=4 and π‘₯=9.

We begin by finding the antiderivative of βˆ’2√π‘₯ from its indefinite integral: ο„Έβˆ’2√π‘₯π‘₯=ο„Έβˆ’2π‘₯π‘₯=βˆ’43π‘₯+𝐢.dd

For the definite integral, we can apply the fundamental theorem of calculus, which states that if 𝑓 is continuous on [π‘Ž,𝑏] and 𝐹′(π‘₯)=𝑓(π‘₯), then 𝑓(π‘₯)π‘₯=[𝐹(π‘₯)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

We note that we can ignore the constant of integration for the antiderivative 𝐹(π‘₯), since this is cancelled in the difference 𝐹(𝑏)βˆ’πΉ(π‘Ž).

The integrand, for our integral, is continuous and defined for all points π‘₯∈[4,9]. Therefore, evaluating the antiderivative at the limits of integration and finding their difference, we have ο„Έβˆ’2√π‘₯π‘₯=ο•βˆ’43π‘₯=ο€½βˆ’43ο€½9ο‰ο‰βˆ’ο€½βˆ’43ο€½4=ο€Όβˆ’43(27)οˆβˆ’ο€Όβˆ’43(8)=βˆ’763.οŠͺοŠͺd

In our next example, we will evaluate the definite integral of an absolute value function.

Example 4: Evaluating the Definite Integral of an Absolute Value Function

Evaluate ο„Έ|π‘₯βˆ’2|π‘₯οŠͺd.

Answer

In this example, we have to evaluate the definite integral of |π‘₯βˆ’2| from π‘₯=βˆ’4 to π‘₯=5. Since the curve, within the interval [βˆ’4,5], lies above the π‘₯-axis, we expect the definite integral to be positive since this gives the signed area under the curve. This is visually represented in the plot, which shows the area under the curve between π‘₯=βˆ’4 and π‘₯=5.

Since the integrand involves a modulus, we recall the definition of |π‘₯|: |π‘₯|=π‘₯π‘₯β‰₯0,βˆ’π‘₯π‘₯<0.forfor

Therefore, for our integrand, we have |π‘₯βˆ’2|=π‘₯βˆ’2π‘₯β‰₯2,2βˆ’π‘₯π‘₯<2.forfor

We begin by finding the antiderivative of |π‘₯βˆ’2| from its indefinite integral, but this will depend on the value of π‘₯. In other words, since |π‘₯βˆ’2| is a piecewise continuous function, we can find the antiderivative of each subfunction piece separately to find the antiderivative of |π‘₯βˆ’2|: ο„Έ|π‘₯βˆ’2|π‘₯=⎧βŽͺ⎨βŽͺβŽ©ο„Έ(π‘₯βˆ’2)π‘₯π‘₯β‰₯2,ο„Έ(2βˆ’π‘₯)π‘₯π‘₯<2,=⎧βŽͺ⎨βŽͺ⎩π‘₯2βˆ’2π‘₯+𝐢π‘₯β‰₯2,2π‘₯βˆ’π‘₯2+𝐢π‘₯<2.ddfordforforfor

For the definite integral, we can apply the fundamental theorem of calculus which states that if 𝑓 is continuous on [π‘Ž,𝑏] and 𝐹′(π‘₯)=𝑓(π‘₯), then 𝑓(π‘₯)π‘₯=[𝐹(π‘₯)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

We note that we can ignore the constant of integration for the antiderivative 𝐹(π‘₯), since this is cancelled in the difference 𝐹(𝑏)βˆ’πΉ(π‘Ž).

In the interval [βˆ’4,5], the definite integral of |π‘₯βˆ’2| will be equivalent to the definite integral of π‘₯βˆ’2 for π‘₯β‰₯2 and the definite integral of 2βˆ’π‘₯ for π‘₯≀2. We can split up the integral into the intervals [βˆ’4,2] and [2,5], so we can perform the integral separately, using the property that if π‘βˆˆ[π‘Ž,𝑏], then 𝑓(π‘₯)π‘₯=𝑓(π‘₯)π‘₯+𝑓(π‘₯)π‘₯.ddd

The integrand, for our integral, is continuous and defined for all points π‘₯∈[βˆ’4,5]. Therefore, evaluating the antiderivative at the limits of integration and finding their difference, we have ο„Έ|π‘₯βˆ’2|π‘₯=ο„Έ|π‘₯βˆ’2|π‘₯+ο„Έ|π‘₯βˆ’2|π‘₯=ο„Έ(2βˆ’π‘₯)π‘₯+ο„Έ(π‘₯βˆ’2)π‘₯=2π‘₯βˆ’π‘₯2+π‘₯2βˆ’2π‘₯=ο€Ύ2(2)βˆ’22οŠβˆ’ο€Ώ2(βˆ’4)βˆ’(βˆ’4)2+ο€Ύ52βˆ’2(5)οŠβˆ’ο€Ύ22βˆ’2(2)=2+ο€Ό8+162+ο€Ό252βˆ’10+2=452.οŠͺοŠͺοŠͺοŠͺddddd

We can also check this answer graphically, since the area under the curve is the sum of the areas of two right triangles, as shown in the graph. We recall that a right triangle with base 𝑏 and height β„Ž has an area of 𝐴=12π‘β„Ž.

The larger triangle has a base and height of 6, while the smaller triangle has a base and height of 3. The definite integral is just the sum of the areas: ο„Έ|π‘₯βˆ’2|π‘₯=12Γ—6Γ—6+12Γ—3Γ—3=362+92=452.οŠͺd

This gives the same result from the fundamental theorem of calculus, as expected.

In our next example, we will evaluate the definite integral of a polynomial.

Example 5: Evaluating the Definite Integral of a Polynomial

Evaluate ο„Έο€Ό45𝑑+34π‘‘βˆ’23π‘‘οˆπ‘‘οŠ§οŠ¦οŠ©οŠ¨d.

Answer

In this example, we have to evaluate the definite integral of 45𝑑+34π‘‘βˆ’23 from 𝑑=0 to 𝑑=1. Since part of the curve, within the interval [0,1], lies above the 𝑑-axis and another part lies below the 𝑑-axis, the definite integral will give the area above 𝑑-axis minus the area under the 𝑑-axis, which we expect to be positive. This is visually represented in the plot, which shows the area under the curve between 𝑑=0 and 𝑑=1.

We begin by finding the antiderivative of 45𝑑+34π‘‘βˆ’23 from its indefinite integral by using the power rule for integration: ο„Έο€Ό45𝑑+34π‘‘βˆ’23π‘‘οˆπ‘‘=15𝑑+14π‘‘βˆ’13𝑑+𝐢.οŠͺd

For the definite integral, we can apply the fundamental theorem of calculus, which states that if 𝑓 is continuous on [π‘Ž,𝑏] and 𝐹′(𝑑)=𝑓(𝑑), then 𝑓(𝑑)𝑑=[𝐹(𝑑)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

We note that we can ignore the constant of integration for the antiderivative 𝐹(𝑑), since this is cancelled in the difference 𝐹(𝑏)βˆ’πΉ(π‘Ž).

The integrand, for our integral, is a polynomial and hence continuous and defined for all points π‘‘βˆˆ[0,1]. Therefore, evaluating the antiderivative at the limits of integration and finding their difference, we have ο„Έο€Ό45𝑑+34π‘‘βˆ’23π‘‘οˆπ‘‘=15𝑑+14π‘‘βˆ’13𝑑=ο€Ό15(1)+14(1)βˆ’13(1)οˆβˆ’ο€Ό15(0)+14(0)βˆ’13(0)=760.οŠͺοŠͺοŠͺd

Since the definite integral gives the signed area under the curve, this is the area in red subtracted from the area in blue. In other words, it is the area under the curve above the 𝑑-axis minus the area under the curve below the 𝑑-axis, in the interval [0,1].

In our next example, we will evaluate the definite integral of an exponential function with an integer base.

Example 6: Evaluating the Definite Integral of an Exponential Function with an Integer Base

Evaluate ο„Έ4𝑠οŠͺd.

Answer

In this example, we have to evaluate the definite integral of 4 from 𝑠=3 to 𝑠=4. Since the curve, within the interval [3,4], lies above the 𝑠-axis, we expect the definite integral to be positive since this gives the signed area under the curve. This is visually represented in the plot, which shows the area under the curve between 𝑠=3 and 𝑠=4.

We begin by finding the antiderivative of 4 from its indefinite integral: ο„Έ4𝑠=𝑒𝑠=14𝑒+𝐢=144+𝐢=1224+𝐢.οŠͺοŠͺddlnlnlnlnln

We have also used the logarithm power rule lnln𝑝=π‘₯𝑝 to rewrite the last line using lnlnln4=2=22. For the definite integral, we can apply the fundamental theorem of calculus, which states that if 𝑓 is continuous on [π‘Ž,𝑏] and 𝐹′(𝑠)=𝑓(𝑠), then 𝑓(𝑠)𝑠=[𝐹(𝑠)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

We note that we can ignore the constant of integration for the antiderivative 𝐹(𝑠), since this is cancelled in the difference 𝐹(𝑏)βˆ’πΉ(π‘Ž).

The integrand, for our integral, is continuous and defined for all points π‘ βˆˆ[3,4]. Therefore, evaluating the antiderivative at the limit of integration and finding their difference, we have ο„Έ4𝑠=1224=ο€Ό1224οˆβˆ’ο€Ό1224=ο€Ό1282οˆβˆ’ο€Ό322=962.οŠͺοŠͺ(οŠͺ)()dlnlnlnlnlnln

In our next example, we will evaluate the definite integral of a power function with a negative fractional exponent.

Example 7: Evaluating the Definite Integral of a Power Function with a Fraction Exponent

Evaluate ο„Έ4π‘₯π‘₯οŽͺd.

Answer

In this example, we have to evaluate the definite integral of 4π‘₯οŽͺ from π‘₯=1 to π‘₯=27. Since the curve, within the interval [1,27], lies above the π‘₯-axis, we expect the definite integral to be positive since this gives the signed area under the curve. This is visually represented in the plot, which shows the area under the curve between π‘₯=1 and π‘₯=27.

We begin by finding the antiderivative of 4π‘₯οŽͺ from its indefinite integral using the power rule for integration: ο„Έ4π‘₯π‘₯=12π‘₯+𝐢.οŽͺd

For the definite integral, we can apply the fundamental theorem of calculus, which states that if 𝑓 is continuous on [π‘Ž,𝑏] and 𝐹′(π‘₯)=𝑓(π‘₯), then 𝑓(π‘₯)π‘₯=[𝐹(π‘₯)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

We note that we can ignore the constant of integration for the antiderivative 𝐹(π‘₯), since this is cancelled in the difference 𝐹(𝑏)βˆ’πΉ(π‘Ž).

The integrand, for our integral, is continuous and defined for all points π‘₯∈[1,27]. Therefore, evaluating the antiderivative at the limits of integration and finding their difference, we have ο„Έ4π‘₯π‘₯=12π‘₯=ο€½12(27)ο‰βˆ’ο€½12(1)=(12(3))βˆ’(12(1))=24.οŽͺd

In our final example, we will evaluate the definite integral of a trigonometric function.

Example 8: Determining the Definite Integral of a Trigonometric Function

Determine ο„Έ(4+πœ‹9π‘₯)π‘₯οŽͺο‘½οŽ₯οŽͺο‘½οŽ£cosd.

Answer

In this example, we have to evaluate the integral of 4+πœ‹9π‘₯cos from π‘₯=βˆ’πœ‹4 to π‘₯=βˆ’πœ‹6. Since the curve, within the interval ο“βˆ’πœ‹4,βˆ’πœ‹6, lies above the π‘₯-axis, we expect the definite integral to be positive since this gives the signed area under the curve. This is visually represented in the plot, which shows the area under the curve between π‘₯=βˆ’πœ‹4 and π‘₯=βˆ’πœ‹6.

We begin by finding the antiderivative of 4+πœ‹9π‘₯cos from its indefinite integral: ο„Έ(4+πœ‹9π‘₯)π‘₯=4π‘₯+πœ‹99π‘₯+𝐢.cosdsin

For the definite integral, we can apply the fundamental theorem of calculus, which states that if 𝑓 is continuous on [π‘Ž,𝑏] and 𝐹′(π‘₯)=𝑓(π‘₯), then 𝑓(π‘₯)π‘₯=[𝐹(π‘₯)]=𝐹(𝑏)βˆ’πΉ(π‘Ž).d

We note that we can ignore the constant of integration for the antiderivative 𝐹(π‘₯), since this is cancelled in the difference 𝐹(𝑏)βˆ’πΉ(π‘Ž).

The integrand, for our integral, is continuous and defined for all points π‘₯βˆˆο“βˆ’πœ‹4,βˆ’πœ‹6. Therefore, evaluating the antiderivative at the limits of integration and finding their difference, we have ο„Έ(4+πœ‹9π‘₯)π‘₯=4π‘₯+πœ‹99π‘₯=4ο€»βˆ’πœ‹6+πœ‹9ο€»9ο€»βˆ’πœ‹6ο‡ο‡οŸβˆ’ο“4ο€»βˆ’πœ‹4+πœ‹9ο€»9ο€»βˆ’πœ‹4ο‡ο‡οŸ=ο”βˆ’2πœ‹3+πœ‹9ο€Όβˆ’9πœ‹6οˆο βˆ’ο”βˆ’πœ‹+πœ‹9ο€Όβˆ’9πœ‹4=ο”βˆ’2πœ‹3βˆ’πœ‹9ο€Ό3πœ‹2οˆο βˆ’ο”βˆ’πœ‹βˆ’πœ‹9ο€Ό9πœ‹4=ο”βˆ’2πœ‹3+πœ‹9ο βˆ’ο—βˆ’πœ‹βˆ’βˆš2πœ‹18=√2πœ‹18+4πœ‹9.οŽͺο‘½οŽ₯οŽͺο‘½οŽ£οŽͺο‘½οŽ₯οŽͺο‘½οŽ£cosdsinsinsinsinsinsinsin

Key Points

  • The definite integral of the continuous function 𝑓(π‘₯) from π‘₯=π‘Ž to π‘₯=𝑏 is the signed area under the curve of 𝑓(π‘₯) from π‘₯=π‘Ž to π‘₯=𝑏.
    The area of the function that lies above the π‘₯-axis in the interval [π‘Ž,𝑏] is positive, while that below the π‘₯-axis is negative.
    If there are parts of the curve that are both above and below the π‘₯-axis in the interval [π‘Ž,𝑏], then the definite integral will be the area above the π‘₯-axis minus the area below the π‘₯-axis.
  • The fundamental theorem of calculus allows us to determine definite integrals from their antiderivative. The corollary to part 1 or part 2 tells us that if 𝑓 is a real-valued continuous function on [π‘Ž,𝑏] and 𝐹 is an antiderivative of 𝑓 (i.e., 𝐹′(π‘₯)=𝑓(π‘₯)) on [π‘Ž,𝑏], then 𝑓(π‘₯)π‘₯=𝐹(𝑏)βˆ’πΉ(π‘Ž).d
  • To evaluate definite integrals in this way, we need to check that 𝑓 is indeed continuous and defined everywhere in the interval [π‘Ž,𝑏].
  • We can ignore the constant of integration or set it to zero for the antiderivatives in definite integrals, as it gets cancelled with the difference 𝐹(𝑏)βˆ’πΉ(π‘Ž).
  • For some piecewise functions or those involving absolute values in the integrand, we may need to split up the integral into multiple parts using the property that if π‘βˆˆ[π‘Ž,𝑏], then 𝑓(π‘₯)π‘₯=𝑓(π‘₯)π‘₯+𝑓(π‘₯)π‘₯.ddd

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