Opening the book…
- Note to the Instructor (click to expand)
- We have arrived—the definite integral as a representation of the precise area under a curve! Beware: Save the reveal of the relationship between the definite integral and *ahem* something else for next section.
- (click to expand)
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- Conic Sections
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The following is a list of learning objectives for this section.
- Learning Objectives (click to expand)
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- State the definition of the definite integral.
- Explain the terms integrand, limits of integration, and variable of integration.
- Explain when a function is integrable.
- Describe the relationship between the definite integral and net area.
- Use Geometry and the properties of definite integrals to evaluate them.
In the preceding section, we defined the area under a curve in terms of Riemann sums:\[A=\lim_{n \to \infty}\sum_{i=1}^nf(x^*_i) \Delta x.\nonumber\]However, this definition came with restrictions. We required \(f(x)\) to be continuous and nonnegative. Unfortunately, real-world problems don't always meet these restrictions. In this section, we look at applying the concept of the area under the curve to a broader set of functions using the definite integral.
Definition and Notation
The definite integral generalizes the concept of the area under a curve. We lift the requirements that \(f(x)\) be continuous and nonnegative, and define the definite integral as follows.
If \(f(x)\) is a function defined on an interval \([a,b]\), the definite integral of \(f\) from \(a\) to \(b\) is given by\[\int^b_af(x)\,dx=\lim_{n \to \infty} \sum_{i=1}^nf(x^*_i) \Delta x,\nonumber\]provided the limit exists. If this limit exists, the function \(f(x)\) is said to be integrable on \([a,b]\), or is an integrable function.
The integral symbol in the previous definition should look familiar. We have seen a similar notation in a previous section, where we used the indefinite integral symbol (without the \(a\) and \(b\) above and below) to represent an antiderivative. Although the notation for indefinite integrals may look similar to that of a definite integral, they are not the same. A definite integral is a number. An indefinite integral is a family of functions. Later in this chapter, we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we're working with a definite integral or an indefinite integral.
Integral notation goes back to the late seventeenth century. It is one of the contributions of Gottfried Wilhelm Leibniz, who is often considered the co-discoverer of Calculus, along with Isaac Newton. The integration symbol \(\int\) is an elongated \(S\), suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of the interval, \([a,b]\). The numbers \(a\) and \(b\) are \(x\)-values and are called the limits of integration; specifically, \(a\) is the lower limit and \(b\) is the upper limit. To clarify, we use the word limit in two different ways in the context of the definite integral. First, we talk about the limit of a sum as \(n \to \infty\). Second, the boundaries of the region are called the limits of integration.
We call the function \(f(x)\) the integrand, and the \(dx\) indicates that \(f(x)\) is a function with respect to \(x\), called the variable of integration. Note that, like the index in a sum, the variable of integration is a dummy variable and has no impact on the computation of the integral. We could use any variable we like as the variable of integration:\[\int^b_af(x)\,dx= \int^b_af(t)\,dt= \int^b_af(u)\, du\nonumber\]Previously, we discussed the fact that if \(f(x)\) is continuous on \([a,b]\), then the limit \(\displaystyle \lim_{n \to \infty}\sum_{i=1}^nf(x^*_i) \Delta x\) exists and is unique. This leads to the following theorem, which we state without proof.
If \(f(x)\) is continuous on \([a,b]\), then \(f\) is integrable on \([a,b]\).
Functions that are not continuous on \([a,b]\) may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities or removable discontinuities on a closed interval are integrable.
It is also worth noting that we are using a regular partition in the Riemann sums. This is not strictly necessary. Any partition can be used to form a Riemann sum. However, if a non-regular partition defines the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero. This introduces a more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.
Use the definition of the definite integral to evaluate \(\displaystyle \int^2_0x^2\,dx\). Use a right-endpoint approximation to generate the Riemann sum.
- Solution
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We first want to set up a Riemann sum. Based on the limits of integration, we have \(a=0\) and \(b=2\). For \(i=0,1,2, \ldots ,n\), let \(P={x_i}\) be a regular partition of \([0,2]\). Then\[\Delta x=\dfrac{b-a}{n}=\dfrac{2}{n}.\nonumber\]Since we are using a right-endpoint approximation to generate Riemann sums, for each \(i\), we need to calculate the function value at the right endpoint of the interval \([x_{i-1},x_i]\). The right endpoint of the interval is \(x_i\), and since \(P\) is a regular partition,\[x_i=x_0+i \Delta x=0+i\left[\dfrac{2}{n}\right]=\dfrac{2i}{n}.\nonumber\]Thus, the function value at the right endpoint of the interval is\[f(x_i)=x^2_i=\left(\dfrac{2i}{n}\right)^2=\dfrac{4i^2}{n^2}.\nonumber\]Then the Riemann sum takes the form\[\sum_{i=1}^nf(x_i) \Delta x=\sum_{i=1}^n\left(\dfrac{4i^2}{n^2}\right)\dfrac{2}{n}=\sum_{i=1}^n\dfrac{8i^2}{n^3}=\dfrac{8}{n^3}\sum_{i=1}^ni^2.\nonumber\]Using the summation formula for \(\displaystyle \sum_{i=1}^ni^2\), we have\[\begin{array}{rcl} \sum_{i=1}^nf(x_i) \Delta x & = & \dfrac{8}{n^3}\sum_{i=1}^ni^2 \\[6pt] & = & \dfrac{8}{n^3}\left[\dfrac{n(n+1)(2n+1)}{6}\right] \\[6pt] & = & \dfrac{8}{n^3}\left[\dfrac{2n^3+3n^2+n}{6}\right] \\[6pt] & = & \dfrac{16n^3+24n^2+8n}{6n^3} \\[6pt] & = & \dfrac{8}{3}+\dfrac{4}{n}+\dfrac{4}{3n^2}. \\[6pt] \end{array}\nonumber\]Now, to calculate the definite integral, we need to take the limit as \(n \to \infty\). We get\[\begin{array}{rcl} \int^2_0x^2dx & = & \lim_{n \to \infty}\sum_{i=1}^nf(x_i) \Delta x \\[6pt] & = & \lim_{n \to \infty}\left(\dfrac{8}{3}+\dfrac{4}{n}+\dfrac{4}{3n^2}\right) \\[6pt] & = & \lim_{n \to \infty}\left(\dfrac{8}{3}\right)+\lim_{n \to \infty}\left(\dfrac{4}{n}\right)+\lim_{n \to \infty}\left(\dfrac{4}{3n^2}\right) \\[6pt] & = & \dfrac{8}{3}+0+0 \\[6pt] & = & \dfrac{8}{3}. \\[6pt] \end{array}\nonumber\]
Use the definition of the definite integral to evaluate \(\displaystyle \int^3_0(2x-1)\,dx\). Use a right-endpoint approximation to generate the Riemann sum.
- Answer
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6
Evaluating Definite Integrals Using Geometry
Evaluating definite integrals this way can be tedious because of the complexity of the calculations. Later in this chapter, we develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we can rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can then discuss what to do in the case of a curve of a function dropping below the \(x\)-axis.
Use the formula for the area of a circle to evaluate \(\displaystyle \int^6_3\sqrt{9-(x-3)^2}\, dx\).
- Solution
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The function describes a semicircle with radius 3. To find\[\int^6_3\sqrt{9-(x-3)^2}\, dx\nonumber\]we want to find the area under the curve over the interval \([3,6]\). The formula for the area of a circle is \(A=\pi r^2\). The area of a semicircle is just one-half the area of a circle, or \(A=\frac{1}{2}\pi r^2\). The shaded area in Figure \(\PageIndex{1}\) covers one-half of the semicircle, or \(A=\frac{1}{4}\pi r^2\). Thus,\[\int^6_3\sqrt{9-(x-3)^2}\,dx=\dfrac{1}{4} \pi (3)^2=\dfrac{9}{4} \pi \approx 7.069.\nonumber\]
![A graph of a semicircle in quadrant one over the interval [0,6] with center at (3,0). The area under the curve over the interval [3,6] is shaded in blue.](https://math.libretexts.org/@api/deki/files/12421/5.2.1.png?revision=1)
Figure \(\PageIndex{1}\): The value of the integral of the function \(f(x)\) over the interval \([3,6]\) is the area of the shaded region.
Use the formula for the area of a trapezoid to evaluate \(\displaystyle \int^4_2(2x+3)\, dx\).
- Answer
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18 square units
The ability to think geometrically can be very useful when it comes to evaluating definite integrals. It's always a good idea to see if the integrand represents a simple geometric curve for which you have an area formula. This will often be circles and lines.
Area and the Definite Integral
When we defined the definite integral, we lifted the requirement that \(f(x)\) be nonnegative. But how do we interpret "the area under the curve" when \(f(x)\) is negative?
Let us return to the Riemann sum. Consider, for example, the function \(f(x)=2-2x^2\) (shown in Figure \(\PageIndex{2}\)) on the interval \([0,2]\). Use \(n=8\) and choose \(x^*_i\) as the left-endpoint of each interval. Construct a rectangle on each subinterval of height \(f(x^*_i)\) and width \(\Delta x\). When \(f(x^*_i)\) is positive, the product \(f(x^*_i) \Delta x\) represents the area of the rectangle, as before. When \(f(x^*_i)\) is negative, however, the product \(f(x^*_i) \Delta x\) represents the negative of the area of the rectangle. The Riemann sum then becomes\[\sum_{i=1}^8f(x^*_i) \Delta x= (\text{Area of rectangles above the }x\text{-axis})-(\text{Area of rectangles below the }x\text{-axis})\nonumber\]
![A graph of a downward opening parabola over [-1, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). Eight rectangles are drawn evenly over [0,2] with heights determined by the value of the function at the left endpoints of each.](https://math.libretexts.org/@api/deki/files/2586/CNX_Calc_Figure_05_02_003.jpeg?revision=1&size=bestfit&width=252&height=300)
Taking the limit as \(n \to \infty\), the Riemann sum approaches the area between the curve above the \(x\)-axis and the \(x\)-axis, less the area between the curve below the \(x\)-axis and the \(x\)-axis, as shown in Figure \(\PageIndex{3}\). Then,\[\int^2_0f(x)\,dx=\lim_{n \to \infty}\sum_{i=1}^nf(c_i) \Delta x=A_1-A_2.\nonumber\]The quantity \(A_1-A_2\) is called the net signed area.
![A graph of a downward opening parabola over [-2, 2] with vertex at (0,2) and x-intercepts at (-1,0) and (1,0). The area in quadrant one under the curve is shaded blue and labeled A1. The area in quadrant four above the curve and to the left of x=2 is shaded blue and labeled A2.](https://math.libretexts.org/@api/deki/files/2587/CNX_Calc_Figure_05_02_002.jpeg?revision=1&size=bestfit&width=279&height=300)
Notice that the net signed area can be positive, negative, or zero. The net signed area is positive if the area above the \(x\)-axis is larger. The net signed area is negative if the area below the \(x\)-axis is larger. The net signed area is zero if the areas above and below the \(x\)-axis are equal.
Find the net signed area between the curve of the function \(f(x)=2x\) and the \(x\)-axis over the interval \([-3,3]\).
- Solution
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The function produces a straight line that forms two triangles: one from \(x=-3\) to \(x=0\) and the other from \(x=0\) to \(x=3\) (Figure \(\PageIndex{4}\)). Using the geometric formula for the area of a triangle, \(A=\frac{1}{2}bh\), the area of triangle \(A_1\), above the axis, is\[A_1=\dfrac{1}{2}3(6)=9,\nonumber\]where \(3\) is the base and \(2(3)=6\) is the height. The area of triangle \(A_2\), below the axis, is\[A_2=\dfrac{1}{2}(3)(6)=9,\nonumber\]where \(3\) is the base and \(6\) is the height. Thus, the net area is\[\displaystyle \int^3_{-3}2x\,dx=A_1-A_2=9-9=0.\nonumber\]
![A graph of an increasing line over [-6, 6] going through the origin and (-3, -6) and (3,6). The area under the line in quadrant one over [0,3] is shaded blue and labeled A1, and the area above the line in quadrant three over [-3,0] is shaded blue and labeled A2.](https://math.libretexts.org/@api/deki/files/12422/5.2.2.png?revision=1)
Figure \(\PageIndex{4}\): The area above the curve and below the \(x\)-axis equals the area below the curve and above the \(x\)-axis. Analysis
If \(A_1\) is the area above the \(x\)-axis and \(A_2\) is the area below the \(x\)-axis, then the net area is \(A_1-A_2\). Since the areas of the two triangles are equal, the net area is zero.
Find the net signed area of \(f(x)=x-2\) over the interval \([0,6]\), illustrated in the following image.
- Answer
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6
The following Interactive Element might help you visualize how the definite integral adds the area above the \(x\)-axis and subtracts the area below it.
Interact: Drag the blue point along the \(x\)-axis to change the upper limit \(b\), and watch the green (above-axis) and red (below-axis) regions grow.
Observation: The definite integral equals the net signed area—the green area minus the red area. It stays negative while the red region is larger, hits exactly zero when the two areas balance (at \(b=4\)), and turns positive once the green region takes over.
One application of the definite integral is finding displacement when given a velocity function. If \(v(t)\) represents the velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position. This is an essential application of the definite integral, and we examine it in more detail later in the chapter. We will look at some basic examples involving constant velocities to understand how this works.
When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar. If a car travels away from its starting position in a straight line at a speed of \(70\) mph for \(2\) hours, then it is \(140\) miles away from its original position (Figure \(\PageIndex{6}\)). Using integral notation, we have\[\int^2_0 70\,dt=140 \,\text{miles}.\nonumber\]
![A graph in quadrant 1 with the x-axis labeled as t (hours) and y-axis labeled as v (mi/hr). The area under the line v(t) = 70 is shaded blue over [0,2].](https://math.libretexts.org/@api/deki/files/2590/CNX_Calc_Figure_05_02_015.jpeg?revision=1&size=bestfit&width=482&height=300)
In the context of displacement, the net signed area allows us to consider direction. If a car travels straight north at \(60\) mph for \(2\) hours, it is \(120\) miles north of its starting position. If the car then turns around and travels south at a speed of \(40\) mph for \(3\) hours, it will be back at its starting position (Figure \(\PageIndex{7}\)). Again, using integral notation, we have\[\int^2_060\,dt+ \int^5_2-40\,dt=120-120=0.\nonumber\]In this case, the displacement is zero.
![A graph in quadrants one and four with the x-axis labeled as t (hours) and the y-axis labeled as v (mi/hr). The first part of the graph is the line v(t) = 60 over [0,2], and the area under the line in quadrant one is shaded. The second part of the graph is the line v(t) = -40 over [2,5], and the area above the line in quadrant four is shaded.](https://math.libretexts.org/@api/deki/files/2591/CNX_Calc_Figure_05_02_016.jpeg?revision=1&size=bestfit&width=314&height=300)
Suppose we want to know how far the car travels, regardless of direction. In this case, we want to know the area between the curve and the \(t\)-axis, regardless of whether that area is above or below the axis. This is called the total area.
Graphically, it is easiest to calculate the total area by adding the areas above and below the axis (rather than subtracting the areas below the axis, as we did with the net signed area). To accomplish this mathematically, we use the absolute value function. Thus, the total distance traveled by the car is\[\int^2_0|60|\,dt+ \int^5_2|-40|\,dt= \int^2_060\,dt+ \int^5_240\,dt=120+120=240.\nonumber\]Bringing these ideas together formally, we state the following definitions.
Let \(f(x)\) be an integrable function defined on an interval \([a,b]\). Let \(A_1\) represent the area between \(f(x)\) and the \(x\)-axis that lies above the axis, and let \(A_2\) represent the area between \(f(x)\) and the \(x\)-axis that lies below the axis. Then, the net signed area between \(f(x)\) and the \(x\)-axis is given by\[\int^b_af(x)\,dx=A_1-A_2.\nonumber\]The total area between \(f(x)\) and the \(x\)-axis is given by\[\int^b_a|f(x)|\,dx=A_1+A_2.\nonumber\]
Find the total area between \(f(x)=x-2\) and the \(x\)-axis over the interval \([0,6]\).
- Solution
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Calculate the \(x\)-intercept as \((2,0)\) (set \(y=0\), solve for \(x\)). To find the total area, take the area below the \(x\)-axis over the subinterval \([0,2]\) and add it to the area above the \(x\)-axis on the subinterval \([2,6]\) (Figure \(\PageIndex{8}\)).
Figure \(\PageIndex{8}\): The total area between the line and the \(x\)-axis over \([0,6]\) is \(A_2\) plus \(A_1\). We have\[\int^6_0|x-2|\,dx=A_2+A_1.\nonumber\]Then, using the formula for the area of a triangle, we obtain\[A_2=\dfrac{1}{2}bh=\dfrac{1}{2} \cdot 2 \cdot 2=2\nonumber\]\[A_1=\dfrac{1}{2}bh=\dfrac{1}{2} \cdot 4 \cdot 4=8.\nonumber\]The total area, then, is\[A_1+A_2=8+2=10\,\text{units}^2.\nonumber\]
Find the total area between the function \(f(x)=2x\) and the \(x\)-axis over the interval \([-3,3]\).
- Answer
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\(18\,\text{units}^2\)
Properties of the Definite Integral
The properties of indefinite integrals also apply to definite integrals. Definite integrals also have properties that relate to the limits of integration. These properties, along with the rules of integration that we examine later in this chapter, help us manipulate expressions to evaluate definite integrals.
- If the limits of integration are the same, the integral is just a vertical line and contains no area. That is,\[\int^a_af(x)\,dx=0.\nonumber\]
- If the limits are reversed, place a negative sign before the integral. That is, \[\int^a_bf(x)\,dx=- \int^b_af(x)\,dx.\nonumber\]
- The integral of a sum is the sum of the integrals. That is, \[\int^b_a[f(x)+g(x)]\,dx= \int^b_af(x)\,dx+ \int^b_ag(x)\,dx.\nonumber\]
- The integral of a difference is the difference of the integrals. That is, \[\int^b_a[f(x)-g(x)]\,dx= \int^b_af(x)\,dx- \int^b_ag(x)\,dx.\nonumber\]
- The integral of the product of a constant and a function is equal to the constant multiplied by the integral of the function. That is, for a constant \(c\),\[\int^b_a cf(x)\,dx=c \int^b_af(x)\,dx.\nonumber\]
- An integral can be split into pieces. That is, \[\int^b_af(x)\,dx= \int^c_af(x)\,dx+ \int^b_cf(x)\,dx.\nonumber\]Although this formula normally applies when \(c\) is between \(a\) and \(b\), the formula holds for all values of \(a\), \(b\), and \(c\), provided \(f(x)\) is integrable on the largest interval.
Use the properties of the definite integral to express the definite integral of \(f(x)=-3x^3+2x+2\) over the interval \([-2,1]\) as the sum of three definite integrals.
- Solution
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Using integral notation, we have \(\displaystyle \int^1_{-2}(-3x^3+2x+2)\,dx\). We apply properties 3. and 5. to get\[\begin{array}{rcl} \int^1_{-2}(-3x^3+2x+2)\,dx & = & \int^1_{-2}-3x^3\,dx+ \int^1_{-2}2x\,dx+ \int^1_{-2}2\,dx\\[6pt] & = & -3 \int^1_{-2}x^3\,dx+2 \int^1_{-2}x\,dx+ \int^1_{-2}2\,dx. \\[6pt] \end{array}\nonumber\]
Use the properties of the definite integral to express the definite integral of \(f(x)=6x^3-4x^2+2x-3\) over the interval \([1,3]\) as the sum of four definite integrals.
- Answer
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\(\displaystyle 6 \int^3_1x^3\,dx-4 \int^3_1x^2\,dx+2 \int^3_1x\,dx- \int^3_13\,dx\)
If it is known that \(\displaystyle \int^8_0f(x)\,dx=10\) and \(\displaystyle \int^5_0f(x)\,dx=5\), find the value of \(\displaystyle \int^8_5f(x)\,dx\).
- Solution
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By property 6,\[\int^b_af(x)\,dx= \int^c_af(x)\,dx+ \int^b_cf(x)\,dx.\nonumber\]Thus,\[\begin{array}{rcl} \int^8_0f(x)\,dx & = & \int^5_0f(x)\,dx+ \int^8_5f(x)\,dx \\[6pt] 10 & = & 5+ \int^8_5f(x)\,dx \\[6pt] 5 & = & \int^8_5f(x)\,dx. \\[6pt] \end{array}\nonumber\]
If it is known that \(\displaystyle \int^5_1f(x)\,dx=-3\) and \(\displaystyle \int^5_2f(x)\,dx=4\), find the value of \(\displaystyle \int^2_1f(x)\,dx\).
- Answer
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\(-7\)
Comparison Properties of Integrals
A picture can sometimes tell us more about a function than the results of computations. Comparing functions by their graphs and algebraic expressions can give new insight into the process of integration. Intuitively, we might say that if a function \(f(x)\) is above another function \(g(x)\), then the area between \(f(x)\) and the \(x\)-axis is greater than the area between \(g(x)\) and the \(x\)-axis. This is true depending on the interval over which the comparison is made. The properties of definite integrals are valid whether \(a<b,\,a=b\), or \(a>b\). The following properties, however, concern only the case \(a \leq b\) and are used when we want to compare the sizes of integrals.
- If \(f(x) \geq 0\) for \(a \leq x \leq b\), then\[\int^b_af(x)\,dx \geq 0.\nonumber\]
- If \(f(x) \geq g(x)\) for \(a \leq x \leq b\), then\[\int^b_af(x)\,dx \geq \int^b_ag(x)\,dx.\nonumber\]
- If \(m\) and \(M\) are constants such that \(m \leq f(x) \leq M\) for \(a \leq x \leq b\), then\[m(b-a) \leq \int^b_af(x)\,dx \leq M(b-a).\nonumber\]
Compare \(f(x)=\sqrt{1+x^2}\) and \(g(x)=\sqrt{1+x}\) over the interval \([0,1]\).
- Solution
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Graphing these functions is necessary to understand how they compare over the interval \([0,1]\). Initially, when graphed on a graphing calculator, \(f(x)\) appears to be above \(g(x)\) everywhere. However, the graphs appear the same on the interval \([0,1]\). We need to zoom in to see that \(g(x)\) is above \(f(x)\) on the given interval. The two functions intersect at \(x=0\) and \(x=1\) (Figure \(\PageIndex{9}\)).
![A graph of the function f(x) = sqrt(1 + x^2) in red and g(x) = sqrt(1 + x) in blue over [-2, 3]. The function f(x) appears above g(x) except over the interval [0,1]. A second, zoomed-in graph shows this interval more clearly.](https://math.libretexts.org/@api/deki/files/12425/5.2.5.png?revision=1)
Figure \(\PageIndex{9a}\): The function \(f(x)\) appears above the function \(g(x)\) except over the interval \([0,1]\).
Figure \(\PageIndex{9b}\): Viewing the same graph with a greater zoom shows this more clearly.We can see from the graph that over the interval \([0,1],\,g(x) \geq f(x)\). Comparing the integrals over the specified interval \([0,1]\), we also see that \(\displaystyle \int^1_0g(x)\,dx \geq \int^1_0f(x)\,dx\) (Figure \(\PageIndex{10}\)). The thin, red-shaded area shows just how much difference there is between these two integrals over the interval \([0,1]\).
![A graph showing the functions f(x) = sqrt(1 + x^2) and g(x) = sqrt(1 + x) over [-3, 3]. The area under g(x) in quadrant one over [0,1] is shaded. The area under g(x) and f(x) is included in this shaded area. The second zoomed-in graph shows more clearly that equality between the functions only holds at the endpoints.](https://math.libretexts.org/@api/deki/files/12426/5.2.6.png?revision=1)
Figure \(\PageIndex{10a}\): The graph shows that over the interval \([0,1],g(x) \geq f(x)\), where equality holds only at the endpoints of the interval.
Figure \(\PageIndex{10b}\): Viewing the same graph with a greater zoom shows this more clearly.
This page titled 4.3: The Definite Integral was last modified on Fri, 02 Oct 2026 23:18:56 GMT and is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman via source content that was edited to the style and standards of the LibreTexts platform.
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