From 166b05cb22540409dd75d68a94474058fe80a215 Mon Sep 17 00:00:00 2001 From: Oscar Levin Date: Thu, 13 Nov 2025 12:25:11 -0700 Subject: [PATCH] add exercises for 4.1; setup rest of 4 --- source/main.ptx | 3 +- source/sec-4-1-velocity-distance.xml | 501 +++------------------- source/sec-4-2-Riemann.xml | 428 +------------------ source/sec-4-3-definite-integral.xml | 526 +----------------------- source/sec-4-4-FTC.xml | 593 +-------------------------- 5 files changed, 102 insertions(+), 1949 deletions(-) diff --git a/source/main.ptx b/source/main.ptx index fecb01c..0c0aafe 100644 --- a/source/main.ptx +++ b/source/main.ptx @@ -10,8 +10,7 @@ - + + + Foundations

- Finally, in the last hour she walked - - D_{[2,3]} = 3 \ \text{miles per hour} \cdot 1 \ \text{hour} = 3 \ \text{miles} - , - so the total distance she traveled is - - D = D_{[0,1.5]} + D_{[1.5,2]} + D_{[2,3]} = 4.5 + 2 + 3 = 9.5 \ \text{miles} - . -

- -

- Since the velocity for 1.5 \lt t \lt 2 is v = -4, - indicating motion in the westward direction, - the person first walked 4.5 miles east, - then 2 miles west, followed by 3 more miles east. - Thus, the total change in her position is - - \text{change in position} = 4.5 - 2 + 3 = 5.5 \ \text{miles} - . -

- -

- It's also valuable to think about our conclusions from a graphical perspective. -

- -
- At left, the velocity function of the person walking; at right, the corresponding position function. - -
- -

- In Figure, - we see how the distances we computed can be viewed as areas: - A_1 = 4.5 comes from multiplying rate times time (3 \cdot 1.5), - as do A_2 and A_3. - But while A_2 is an area - (and is therefore positive), - because the velocity function is negative for 1.5 \lt t \lt 2, - this area has a negative sign associated with it. - The negative area distinguishes between distance traveled and change in position. -

- -

- The distance traveled is the sum of the areas, - - D = A_1 + A_2 + A_3 = 4.5 + 2 + 3 = 9.5 \ \text{miles} - . -

- -

- But the change in position has to account for travel in the negative direction. - An area above the t-axis is considered positive because it represents distance traveled in the positive direction, - while one below the t-axis is viewed as negative because it represents travel in thenegative direction. - Thus, the change in the person's position is - - s(3) - s(0) = (+4.5) + (-2) + (+3) = 5.5 \ \text{miles} - . - In other words, - the person walks 4.5 miles in the positive direction, - followed by two miles in the negative direction, - and then 3 more miles in the positive direction. -

- -

- Negative velocity is also seen in the graph of the position function y=s(t). - Its slope is negative (specifically, - -4) on the interval - 1.5\lt t\lt 2 because the velocity is -4 on that interval. - The negative slope shows the position function is decreasing because the person is walking east, - rather than west. -

- -

- To summarize, we see that if velocity is sometimes negative, - a moving object's change in position is different from its distance traveled. - If we compute separately the distance traveled on each interval where velocity is positive or negative, - we can calculate either the total distance traveled or the total change in position. - We assign a negative value to distances traveled in the negative direction when we calculate change in position, - but a positive value when we calculate the total distance traveled. -

- - - - - - Summary -

-

    -
  • -

    - If we know the velocity of a moving body at every point in a given interval and the velocity is positive throughout, - we can estimate the object's distance traveled and in some circumstances determine this value exactly. -

    -
    • - -
  • -

    - In particular, when velocity is positive on an interval, - we can find the total distance traveled by finding the area under the velocity curve and above the t-axis on the given time interval. - We may only be able to estimate this area, - depending on the shape of the velocity curve. -

    -
    • - -
  • -

    - An antiderivative of a function f is a new function F whose derivative is f. - That is, - F is an antiderivative of f provided that F' = f. - In the context of velocity and position, - if we know a velocity function v, - an antiderivative of v is a position function s that satisfies s' = v. - If v is positive on a given interval, - say [a,b], then the change in position, s(b) - s(a), - measures the distance the moving object traveled on [a,b]. -

    -
    • - -
  • -

    - If its velocity is sometimes negative, - a moving object is sometimes traveling in the opposite direction - or backtracking. - To determine distance traveled, - we have to compute the distance separately on intervals where velocity is positive or negative, - and account for the change in position on each such interval. -

    -
    • -
+ Coming soon.

- + + Calculus Practice + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/source/sec-4-2-Riemann.xml b/source/sec-4-2-Riemann.xml index 80a2a02..90efbde 100644 --- a/source/sec-4-2-Riemann.xml +++ b/source/sec-4-2-Riemann.xml @@ -3,7 +3,7 @@
Riemann sums - + + + Foundations

- In Section, - we learned that if an object moves with positive velocity v, - the area between y = v(t) and the t-axis over a given time interval tells us the distance traveled by the object over that time period. - If v(t) is sometimes negative and we view the area of any region below the t-axis as having an associated negative sign, - then the sum of these signed areas tells us the moving object's change in position over a given time interval. + Coming soon.

+ - + + Calculus Practice +

- For instance, - for the velocity function given in Figure, - if the areas of shaded regions are A_1, - A_2, and A_3 as labeled, - then the total distance D traveled by the moving object on [a,b] is - - D = A_1 + A_2 + A_3 - , - while the total change in the object's position on [a,b] is - - s(b) - s(a) = A_1 - A_2 + A_3 - . - Because the motion is in the negative direction on the interval where v(t) \lt 0, - we subtract A_2 to determine the object's total change in position. + Coming soon.

- -
- A velocity function that is sometimes negative. - -
- - -

- Of course, finding D and - s(b)-s(a) for the graph in Figure - presumes that we can actually find the areas A_1, - A_2, and A_3. - So far, - we have worked with velocity functions that were either constant or linear - so that we could find the exact area bounded by the velocity function and the horizontal axis through a combination of rectangles and triangles. - But when the curve bounds a region that is not a familiar geometric shape, - we cannot find its area exactly without some key ideas in calculus. - Indeed, this is one of our biggest goals in Chapter: - to learn how to find the exact area bounded between a curve and the horizontal axis for as many different types of functions as possible. -

- -

- In Activity, - we approximated the area under a nonlinear velocity function using rectangles. - In the following preview activity, - we consider three different options for the heights of the rectangles we will use. -

- - - - - - - - - Sigma Notation -

- We have used sums of areas of rectangles to approximate the area under a curve. - Intuitively, - we expect that using a larger number of thinner rectangles will provide a better estimate for the area. - Consequently, - we anticipate dealing with sums of a large number of terms. - To do so, we introduce sigma notation, - sigma notation - named for the Greek letter \Sigma, - which is the capital letter S in the Greek alphabet. -

- -

- For example, say we are interested in the sum - - 1 + 2 + 3 + \cdots + 100 - , - the sum of the first 100 natural numbers. - In sigma notation we write - - \sum_{k=1}^{100} k = 1 + 2 + 3 + \cdots + 100 - . -

- -

- We read the symbol \sum_{k=1}^{100} k as - the sum from k equals 1 to 100 of k. - The variable k is called the index of summation, - and any letter can be used for this variable. - The pattern in the terms of the sum is denoted by a function of the index; - for example, - - \sum_{k=1}^{10} (k^2 + 2k) = (1^2 + 2\cdot 1) + (2^2 + 2\cdot 2) + (3^2 + 2\cdot 3) + \cdots + (10^2 + 2\cdot 10) - , - and more generally, - - \sum_{k=1}^n f(k) = f(1) + f(2) + \cdots + f(n) - . -

- -

- Sigma notation allows us to easily change the function being used to describe the terms in the sum, - and to adjust the number of terms in the sum simply by changing the value of n. - We test our understanding of this new notation in the following activity. -

- - - - - - Riemann Sums -

- When a moving body has a positive velocity function - y = v(t) on a given interval [a,b], - the area under the curve over the interval gives the total distance the body travels on [a,b]. - We are also interested in finding the exact area bounded by - y = f(x) on an interval [a,b], - regardless of the meaning or context of the function f. - For now, we continue to focus on finding an accurate estimate of this area by using a sum of the areas of rectangles. - Unless otherwise indicated, - we assume that f is continuous and non-negative on [a,b]. -

- -

- The first choice we make in such an approximation is the number of rectangles. -

- -
- Subdividing the interval [a,b] into n subintervals of equal length \Delta x. - -
- -

- If we desire n rectangles of equal width to subdivide the interval [a,b], - then each rectangle must have width \Delta x = \frac{b-a}{n}. - We let x_0 = a, x_n = b, - and define x_{i} = a + i\Delta x, - so that x_1 = x_0 + \Delta x, - x_2 = x_0 + 2 \Delta x, - and so on, as pictured in Figure. -

- -

- We use each subinterval [x_i, x_{i+1}] as the base of a rectangle, - and next choose the height of the rectangle on that subinterval. - There are three standard choices: we can - use the left endpoint of each subinterval, - the right endpoint of each subinterval, or the midpoint of each. - These are precisely the options encountered in Preview Activity - and seen in its figure. - We next explore how these choices can be described in sigma notation. -

- -

- Consider an arbitrary positive function f on [a,b] with the interval subdivided as shown in Figure, - and choose to use left endpoints. - Then on each interval [x_{i}, x_{i+1}], - the area of the rectangle formed is given by - - A_{i+1} = f(x_i) \cdot \Delta x - , - as seen in Figure. -

- -

- If we let L_n denote the sum of the areas of these rectangles, - we see that - - L_n =\mathstrut \amp A_1 + A_2 + \cdots + A_{i+1} + \cdots + A_n - =\mathstrut \amp f(x_0) \cdot \Delta x + f(x_1) \cdot \Delta x + \cdots + f(x_i) \cdot \Delta x + \cdots + f(x_{n-1}) \cdot \Delta x - . - In the more compact sigma notation, we have - - L_n = \sum_{i = 0}^{n-1} f(x_i) \Delta x - . -

- -
- Subdividing the interval [a,b] into n subintervals of equal length \Delta x and approximating the area under y = f(x) over [a,b] using left rectangles. - -
- -

- Note that since the index of summation begins at 0 and ends at n-1, - there are indeed n terms in this sum. - We call L_n the left Riemann sum - Riemann sum - Riemann sumleft - for the function f on the interval [a,b]. -

- -

- To see how the Riemann sums for right endpoints and midpoints are constructed, - we consider Figure. -

- -
- Riemann sums using right endpoints and midpoints. - -
- -

- For the sum with right endpoints, - we see that the area of the rectangle on an arbitrary interval - [x_i, x_{i+1}] is given by B_{i+1} = f(x_{i+1}) \cdot \Delta x, - and that the sum of all such areas of rectangles is given by - - R_n =\mathstrut \amp B_1 + B_2 + \cdots + B_{i+1} + \cdots + B_n - =\mathstrut \amp f(x_1) \cdot \Delta x + f(x_2) \cdot \Delta x + \cdots + f(x_{i+1}) \cdot \Delta x + \cdots + f(x_{n}) \cdot \Delta x - =\mathstrut \amp \sum_{i=1}^{n} f(x_i) \Delta x - . -

- -

- We call R_n the right Riemann sum - Riemann sumright - for the function f on the interval [a,b]. -

- -

- For the sum that uses midpoints, we introduce the notation - - \overline{x}_{i+1} = \frac{x_{i} + x_{i+1}}{2} - - so that \overline{x}_{i+1} is the midpoint of the interval [x_i, x_{i+1}]. - For instance, - for the rectangle with area C_1 in Figure, - we now have - - C_1 = f(\overline{x}_1) \cdot \Delta x - . -

- -

- Hence, the sum of all the areas of rectangles that use midpoints is - - M_n =\mathstrut \amp C_1 + C_2 + \cdots + C_{i+1} + \cdots + C_n - =\mathstrut \amp f(\overline{x}_1) \cdot \Delta x + f(\overline{x}_2) \cdot \Delta x + \cdots + f(\overline{x}_{i+1}) \cdot \Delta x + \cdots + f(\overline{x}_{n}) \cdot \Delta x - =\mathstrut \amp \sum_{i=1}^{n} f(\overline{x}_i) \Delta x - , - and we say that M_n is the - middle Riemann sum - Riemann summiddle - for f on [a,b]. -

- -

- Thus, we have two variables to explore: - the number of rectangles and the height of each rectangle. - We can explore these choices dynamically, - and this interactive - is a particularly useful one. - There we see the image shown in Figure, - but with the opportunity to adjust the slider bars for the heights and the number of rectangles. -

- -
- A snapshot of the interactive graphic found at gvsu.edu/s/a9. - -
- -

- By moving the sliders, - we can see how the heights of the rectangles change as we consider left endpoints, - midpoints, and right endpoints, - as well as the impact that a larger number of narrower rectangles has on the approximation of the exact area bounded by the function and the horizontal axis. -

- -

- When f(x) \ge 0 on [a,b], - each of the Riemann sums L_n, R_n, - and M_n provides an estimate of the area under the curve - y = f(x) over the interval [a,b]. - We also recall that in the context of a nonnegative velocity function y = v(t), - the corresponding Riemann sums approximate the distance traveled on [a,b] by a moving object with velocity function v. -

- -

- There is a more general way to think of Riemann sums, - and that is to allow any choice of where the function is evaluated to determine the rectangle heights. - Rather than saying we'll always choose left endpoints, - or always choose midpoints, - we simply say that a point - x_{i+1}^* will be selected at random in the interval [x_i, x_{i+1}] - (so that x_i \le x_{i+1}^* \le x_{i+1}). - The Riemann sum is then given by - - f(x_1^*) \cdot \Delta x + f(x_2^*) \cdot \Delta x + \cdots + f(x_{i+1}^*) \cdot \Delta x + \cdots + f(x_n^*) \cdot \Delta x = \sum_{i=1}^{n} f(x_i^*) \Delta x - . -

- -

- - In the interactive graphic noted earlier - and referenced in Figure, - by unchecking the relative box at the top left, - and instead checking random, - we can easily explore the effect of using random point locations in subintervals on a Riemann sum. - In computational practice, we most often use L_n, - R_n, or M_n, - while the random Riemann sum is useful in theoretical discussions. - In the following activity, - we investigate several different Riemann sums for a particular velocity function. -

- - - - - - When the function is sometimes negative -

- For a Riemann sum such as - - L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x - , - we can of course compute the sum even when f takes on negative values. - We know that when f is positive on [a,b], - a Riemann sum estimates the area bounded between f and the horizontal axis over the interval. -

- -
- At left and center, two left Riemann sums for a function f that is sometimes negative; at right, the areas bounded by f on the interval [a,d]. - -
- -

- For the function pictured in the first graph of Figure, - a left Riemann sum with 12 subintervals over [a,d] is shown. - The function is negative on the interval b \le x \le c, - so at the four left endpoints that fall in [b,c], - the terms f(x_i) \Delta x are negative. - This means that those four terms in the Riemann sum produce an estimate of the opposite - of the area bounded by y = f(x) and the x-axis on [b,c]. -

- -

- In the middle graph of Figure, - we see that increasing the number of rectangles appears to improve the approximation of the area - (or the opposite of the area) - bounded by the curve. -

- -

- In general, - any Riemann sum of a continuous function f on an interval [a,b] approximates the difference between the area that lies above the horizontal axis on [a,b] and under f and the area that lies below the horizontal axis on [a,b] and above f. - In the notation of Figure, - we may say that - - L_{24} \approx A_1 - A_2 + A_3 - , - where L_{24} is the left Riemann sum using 24 subintervals shown in the middle graph. - A_1 and A_3 are the areas of the regions where f is positive, - and A_2 is the area where f is negative. - We will call the quantity A_1 - A_2 + A_3 the - net signed area - net signed area - bounded by f over the interval [a,d], - where by the phrase signed area - we indicate that we are attaching a minus sign to the areas of regions that fall below the horizontal axis. -

- -

- Finally, we recall that if the function f represents the velocity of a moving object, - the sum of the areas bounded by the curve tells us the total distance traveled over the relevant time interval, - while the net signed area bounded by the curve computes the object's change in position on the interval. -

- - - - - - Summary -

-

    -
  • -

    - A Riemann sum is simply a sum of products of the form - f(x_i^*) \Delta x that estimates the area between a positive function and the horizontal axis over a given interval. - If the function is sometimes negative on the interval, - the Riemann sum estimates the difference between the areas that lie above the horizontal axis and those that lie below the axis. -

    -
    • - -
  • -

    - The three most common types of Riemann sums are left, right, - and middle sums, - but we can also work with a more general Riemann sum. - The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed. - For a left Riemann sum, - we evaluate the function at the left endpoint of each subinterval, - while for right and middle sums, - we use right endpoints and midpoints, respectively. -

    -
    • - -
  • -

    - The left, - right, and middle Riemann sums are denoted L_n, - R_n, and M_n, with formulas - - L_n = f(x_0) \Delta x + f(x_1) \Delta x + \cdots + f(x_{n-1}) \Delta x \amp= \sum_{i = 0}^{n-1} f(x_i) \Delta x, - R_n = f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(x_i) \Delta x, - M_n = f(\overline{x}_1) \Delta x + f(\overline{x}_2) \Delta x + \cdots + f(\overline{x}_{n}) \Delta x \amp= \sum_{i = 1}^{n} f(\overline{x}_i) \Delta x - , - where x_0 = a, x_i = a + i\Delta x, - and x_n = b, using \Delta x = \frac{b-a}{n}. - For the midpoint sum, \overline{x}_{i} = (x_{i-1} + x_i)/2. -

    -
    • -
-

- - - + +
diff --git a/source/sec-4-3-definite-integral.xml b/source/sec-4-3-definite-integral.xml index 7b3d415..df07919 100644 --- a/source/sec-4-3-definite-integral.xml +++ b/source/sec-4-3-definite-integral.xml @@ -3,7 +3,7 @@
The definite integral - + -

- For example, - if f is the function pictured in Figure, - and A_1, - A_2, - and A_3 are the exact areas bounded by f and the x-axis on the respective intervals [a,b], - [b,c], and [c,d], then - - \int_a^b f(x) \, dx = A_1, \ \int_b^c f(x) \, dx = -A_2, - \int_c^d f(x) \, dx = A_3, - \text{ and } \int_a^d f(x) \, dx = A_1 - A_2 + A_3 - . -

- -
- A continuous function f on the interval [a,d]. - -
- -

- We can also use definite integrals to express the change in position and the distance traveled by a moving object. - If v is a velocity function on an interval [a,b], - then the change in position of the object, - s(b) - s(a), is given by - - s(b) - s(a) = \int_a^b v(t) \, dt - . -

- -

- If the velocity function is nonnegative on [a,b], - then \int_a^b v(t) \,dt tells us the distance the object traveled. - If the velocity is sometimes negative on [a,b], - we can use definite integrals to find the areas bounded by the function on each interval where v does not change sign, - and the sum of these areas will tell us the distance the object traveled. -

- -

- To compute the value of a definite integral from the definition, - we have to take the limit of a sum. - While this is possible to do in select circumstances, - it is also tedious and time-consuming, - and does not offer much additional insight into the meaning or interpretation of the definite integral. - Instead, in Section, - we will learn the Fundamental Theorem of Calculus, - which provides a shortcut for evaluating a large class of definite integrals. - This will enable us to determine the exact net signed area bounded by a continuous function and the x-axis in many circumstances. -

- -

- For now, our goal is to understand the meaning and properties of the definite integral, - rather than to compute its value. - To do this, we will rely on the net signed area interpretation of the definite integral. - So we will use as examples curves that produce regions whose areas we can compute exactly through area formulas. - We can thus compute the exact value of the corresponding integral. -

- -

- For instance, - if we wish to evaluate the definite integral \int_1^4 (2x+1) \, dx, - we observe that the region bounded by this function and the x-axis is the trapezoid shown in Figure. - By the formula for the area of a trapezoid, - A = \frac{1}{2}(3+9) \cdot 3 = 18, so - - \int_1^4 (2x+1) \, dx = 18 - . -

- -
- The area bounded by f(x)=2x+1 and the x-axis on the interval [1,4]. - -
- - - - - - Some properties of the definite integral -

- Regarding the definite integral of a function f over an interval [a,b] as the net signed area bounded by f and the x-axis, - we discover several standard properties of the definite integral. - It is helpful to remember that the definite integral is defined in terms of Riemann sums, - which consist of the areas of rectangles. -

- -

- For any real number a and the definite integral - \int_a^a f(x) \, dx - it is evident that no area is enclosed, - because the interval begins and ends with the same point. - Hence, -

- - - Integrating from <m>x = a</m> to <m>x = a</m> - -

- If f is a continuous function and a is a real number, - then \int_a^a f(x) \,dx = 0. -

- - -

- Next, we consider the result of subdividing the interval of integration. - In Figure, we see that - - \int_a^b f(x) \, dx = A_1, \ \int_b^c f(x) \, dx = A_2, - \text{and }\int_a^c f(x) \, dx = A_1 + A_2 - , - which illustrates the following general rule. -

- -
- The area bounded by y=f(x) on the interval [a,c]. - -
- - - Subdividing the interval of integration - -

- If f is a continuous function and a, b, - and c are real numbers, then - - \int_a^c f(x) \,dx = \int_a^b f(x) \,dx + \int_b^c f(x) \,dx - . -

- - -

- While this rule is easy to see if a \lt b \lt c, - it in fact holds in general for any values of a, - b, - and c. - Another property of the definite integral states that if we reverse the order of the limits of integration, - we change the sign of the integral's value. -

- - - Reversing the limits of integration - -

- If f is a continuous function and a and b are real numbers, then - - \int_b^a f(x) \,dx = -\int_a^b f(x) \,dx - . -

- - -

- This result makes sense because if we integrate from a to b, - then in the defining Riemann sum we set \Delta x = \frac{b-a}{n}, - while if we integrate from b to a, we have - \Delta x = \frac{a-b}{n} = -\frac{b-a}{n}, - and this is the only change in the sum used to define the integral. -

- + --> + + Foundations

- There are two additional useful properties of the definite integral. - When we worked with derivative rules in Chapter, - we learned the Constant Multiple Rule and the Sum Rule. - Recall that the Constant Multiple Rule says that if f is a differentiable function and k is a constant, then - - \frac{d}{dx} [kf(x)] = kf'(x) - , - and the Sum Rule says that if f and g are differentiable functions, then - - \frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x) - . + Coming soon.

- -

- These rules are useful because they allow to deal individually with the simplest parts of certain functions by taking advantage of addition and multiplying by a constant. - In other words, - the process of taking the derivative respects addition and multiplying by constants in the simplest possible way. -

- -

- It turns out that similar rules hold for the definite integral. - First, let's consider the functions pictured in Figure. -

- -
- The areas bounded by y = f(x) and y = 2f(x) on [a,b]. - -
- -

- Because multiplying the function by 2 doubles its height at every x-value, - we see that the height of each rectangle in a left Riemann sum is doubled, - f(x_i) for the original function, - versus 2f(x_i) in the doubled function. - For the areas A and B, - it follows B = 2A. - As this is true - regardless of the value of n or the type of sum we use, - we see that in the limit, - the area of the red region bounded by - y = 2f(x) will be twice the area of the blue region bounded by y = f(x). - As there is nothing special about the value 2 compared to an arbitrary constant k, - the following general principle holds. -

- - - The Constant Multiple Rule - definite integralconstant multiple rule - -

- If f is a continuous function and k is any real number, then - - \int_a^b k \cdot f(x) \,dx = k \int_a^b f(x) \,dx - . -

- - -

- We see a similar situation with the sum of two functions f and g. -

- -
- The areas bounded by y = f(x) and y = g(x) on [a,b], as well as the area bounded by y = f(x) + g(x). - -
- -

- If we take the sum of two functions f and g - at every point in the interval, - the height of the function f+g is given by (f+g)(x_i) = f(x_i) + g(x_i). - Hence, for the pictured rectangles with areas A, - B, - and C, - it follows that C = A + B. - Because this will occur for every such rectangle, - in the limit the area of the gray region will be the sum of the areas of the blue and red regions. - In terms of definite integrals, - we have the following general rule. -

- - - The Sum Rule - definite integralsum rule - -

- If f and g are continuous functions, then - - \int_a^b [f(x) + g(x)] \,dx = \int_a^b f(x) \,dx + \int_a^b g(x) \,dx - . -

- - -

- The Constant Multiple and Sum Rules can be combined to say that for any continuous functions f and g and any constants c and k, - - \int_a^b [c f(x) \pm k g(x)] \,dx = c \int_a^b f(x) \,dx \pm k \int_a^b g(x) \,dx - . -

- - - - How the definite integral is connected to a function's average value - average value of a function - -

- One of the most valuable applications of the definite integral is that it provides a way to discuss the average value of a function, - even for a function that takes on infinitely many values. - Recall that if we wish to take the average of n numbers y_1, - y_2, \ldots, y_n, we compute - - AVG = \frac{y_1 + y_2 + \cdots + y_n}{n} - . -

- -

- Since integrals arise from Riemann sums in which we add n values of a function, - it should not be surprising that evaluating an integral is similar to averaging the output values of a function. - Consider, for instance, - the right Riemann sum R_n of a function f, - which is given by - - R_n = f(x_1) \Delta x + f(x_2) \Delta x + \cdots + f(x_n) \Delta x = (f(x_1) + f(x_2) + \cdots + f(x_n))\Delta x - . -

- -

- Since \Delta x = \frac{b-a}{n}, we can thus write - - R_n =\mathstrut \amp (f(x_1) + f(x_2) + \cdots + f(x_n))\cdot \frac{b-a}{n} - =\mathstrut \amp (b-a) \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n} - . -

- -

- We see that the right Riemann sum with n subintervals is just the length of the interval (b-a) times the average of the n function values found at the right endpoints. - And just as with our efforts to compute area, - the larger the value of n we use, - the more accurate our average will be. - Indeed, we will define the average value of f on [a,b] to be - - f_{\operatorname{AVG} [a,b]} = \lim_{n \to \infty} \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n} - . -

- -

- But we also know that for any continuous function f on [a,b], - taking the limit of a Riemann sum leads precisely to the definite integral. - That is, \lim_{n \to \infty} R_n = \int_a^b f(x) \, dx, - and thus taking the limit as - n \to \infty in Equation, - we have that - - \int_a^b f(x) \, dx = (b-a) \cdot f_{\operatorname{AVG} [a,b]} - . -

- -

- Solving Equation for f_{\operatorname{AVG} [a,b]}, - we have the following general principle. -

- - - The average value of a function + + Calculus Practice +

- average value - If f is a continuous function on [a,b], - then its average value on [a,b] is given by the formula - - f_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \cdot \int_a^b f(x) \, dx - . + Coming soon.

- - -

- Equation tells us another way to interpret the definite integral: - the definite integral of a function f from a to b is the length of the interval (b-a) times the average value of the function on the interval. - In addition, - when the function f is nonnegative on [a,b], Equation has a natural visual interpretation. -

- -
- A function y = f(x), the area it bounds, and its average value on [a,b]. - -
- -

- Consider Figure, - where we see at left the shaded region whose area is \int_a^b f(x) \, dx, - at center the shaded rectangle whose dimensions are (b-a) by f_{\operatorname{AVG} [a,b]}, - and at right these two figures superimposed. - Note that in dark green we show the horizontal line y = f_{\operatorname{AVG} [a,b]}. - Thus, the area of the green rectangle is given by (b-a) \cdot f_{\operatorname{AVG} [a,b]}, - which is precisely the value of \int_a^b f(x) \, dx. - The area of the blue region in the left figure is the same as the area of the green rectangle in the center figure. - We can also observe that the areas A_1 and A_2 in the rightmost figure appear to be equal. - Thus, knowing the average value of a function enables us to construct a rectangle whose area is the same as the value of the definite integral of the function on the interval. - This interactive graphic - provides an opportunity to explore how the average value of the function changes as the interval changes, - through an image similar to that found in Figure. -

- - - - - - Summary -

-

    -
  • -

    - Any Riemann sum of a continuous function f on an interval [a,b] provides an estimate of the net signed area bounded by the function and the horizontal axis on the interval. - Increasing the number of subintervals in the Riemann sum improves the accuracy of this estimate, - and letting the number of subintervals increase without bound results in the values of the corresponding Riemann sums approaching the exact value of the enclosed net signed area. -

    -
    • - -
  • -

    - When we take the limit of Riemann sums, - we arrive at what we call the definite integral of f over the interval [a,b]. - In particular, the symbol - \int_a^b f(x) \, dx denotes the definite integral of f over [a,b], - and this quantity is defined by the equation - - \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x - , - where \Delta x = \frac{b-a}{n}, - x_i = a + i\Delta x - (for i = 0, \ldots, n), - and x_i^* satisfies - x_{i-1} \le x_i^* \le x_i - (for i = 1, \ldots, n). -

    -
    • - -
  • -

    - The definite integral - \int_a^b f(x) \,dx measures the exact net signed area bounded by f and the horizontal axis on [a,b]; - in addition, - the value of the definite integral is related to what we call the average value of the function on [a,b]: - f_{\text{AVG} [a,b]} = \frac{1}{b-a} \cdot \int_a^b f(x) \, dx. - In the setting where we consider the integral of a velocity function v, - \int_a^b v(t) \,dt measures the exact change in position of the moving object on [a,b]; - when v is nonnegative, - \int_a^b v(t) \,dt is the object's distance traveled on [a,b]. -

    -
    • - -
  • -

    - The definite integral is a sophisticated sum, - and thus has some of the same natural properties that finite sums have. - Perhaps most important of these is how the definite integral respects sums and constant multiples of functions, - which can be summarized by the rule - - \int_a^b [c f(x) \pm k g(x)] \,dx = c \int_a^b f(x) \,dx \pm k \int_a^b g(x) \,dx - - where f and g are continuous functions on [a,b] and c and k are arbitrary constants. -

    -
    • -
-

- - - + +
diff --git a/source/sec-4-4-FTC.xml b/source/sec-4-4-FTC.xml index eea8c00..1dc2c0d 100644 --- a/source/sec-4-4-FTC.xml +++ b/source/sec-4-4-FTC.xml @@ -3,7 +3,7 @@
The Fundamental Theorem of Calculus - + + + Foundations

- Since F(x) = \ln(|x|) is an antiderivative of f(x) = \frac{1}{x} for all x \ne 0, it seems natural to write that the general antiderivative of f(x) = \frac{1}{x} is F(x) = \ln(|x|) + C. But doing so is technically incorrect. This is due to the discontinuity of f(x) = \frac{1}{x} at x = 0 and the fact that there's one antiderivative, \ln(-x) + C_1, for x \lt 0, and a second, \ln(x) + C_2, for x \gt 0. While we will still sometimes write F(x) = \ln(|x|) + C as the general antiderivative, we should be especially careful when the value x = 0, where \frac{1}{x} is undefined, is included in the interval of values in which we are interested. + Coming soon.

-

- This situation arises for any function that has a discontinuity (e.g., \frac{1}{x^2}, \tan(x)); some of these issues will be considered further when we study improper integrals in Section. -

- - - The total change theorem - total change theorem - -

- Let us review three interpretations of the definite integral. - -

    -
  • -

    - For a moving object with instantaneous velocity v(t), - the object's change in position on the time interval [a,b] is given by \int_a^b v(t) \, dt, - and whenever v(t) \ge 0 on [a,b], - \int_a^b v(t) \, dt tells us the total distance traveled by the object on [a,b]. -

    -
    • - -
  • -

    - For any continuous function f, - its definite integral - \int_a^b f(x) \, dx represents the net signed area bounded by - y = f(x) and the x-axis on [a,b], - where regions that lie below the x-axis have a minus sign associated with their area. -

    -
    • - -
  • -

    - The value of a definite integral is linked to the average value of a function: - for a continuous function f on [a,b], - its average value f_{\operatorname{AVG} [a,b]} is given by - - f_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \int_a^b f(x) \, dx - . -

    -
    • -
-

- -

- The Fundamental Theorem of Calculus now enables us to evaluate exactly - (without taking a limit of Riemann sums) - any definite integral for which we are able to find an antiderivative of the integrand. -

- -

- A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. - Recall Equation, - where we wrote the Fundamental Theorem of Calculus for a velocity function v with antiderivative V as - - V(b) - V(a) = \int_a^b v(t) \, dt - . -

- -

- If we instead replace V with s - (which represents position) - and replace v with s' - (since velocity is the derivative of position), - Equation then reads as - - s(b) - s(a) = \int_a^b s'(t) \, dt - . -

- -

- In words, this version of the FTC tells us that the total change in an object's position function on a particular interval is given by the definite integral of the position function's derivative over that interval. -

- -

- Of course, this result is not limited to only the setting of position and velocity. - Writing the result in terms of a more general function f, - we have the Total Change Theorem. -

- - - The Total Change Theorem + + Calculus Practice +

- total change theorem - If f is a continuously differentiable function on [a,b] with derivative f', - then f(b) - f(a) = \int_a^b f'(x) \, dx. - That is, the definite integral of the rate of change of a function on [a,b] is the total change of the function itself on [a,b]. + Coming soon.

- - -

- The Total Change Theorem tells us more about the relationship between the graph of a function and that of its derivative. - Recall that heights on the graph of the derivative function are equal to slopes on the graph of the function itself. - If instead we know f' and are seeking information about f, - we can say the following: -

- -
-

- differences in heights on f correspond to net signed areas bounded by f'. -

-
- -

- To see why this is so, let's revisit the function f(x) = 4x - x^2 and its derivative f'(x) = 4 - 2x that were the focus in Figure. As one example of a difference in heights on the graph of f, consider the difference f(1) - f(0). - This value is 3, because f(1) = 3 and f(0) = 0, - but also because the net signed area bounded by - y = f'(x) on [0,1] is 3. - That is, - - f(1) - f(0) = \int_0^1 f'(x) \, dx - . - Again, this means that differences in heights on f correspond to net signed areas bounded by f', as shown in Figure. -

- -
- The graphs of f'(x) = 4 - 2x (at left) and an antiderivative f(x) = 4x - x^2 at right. Differences in heights on f correspond to net signed areas bounded by f'. - -
- -

- In addition to this observation about area, - the Total Change Theorem enables us to answer questions about a function whose rate of change we know. -

- - - -

- Suppose that pollutants are leaking out of an underground storage tank at a rate of r(t) gallons/day, - where t is measured in days. - It is conjectured that r(t) is given by the formula - r(t) = 0.0069t^3 -0.125t^2+11.079 over a certain 12-day period. - What is the meaning of \int_4^{10} r(t) \, dt and what is its value? - What is the average rate at which pollutants are leaving the tank on the time interval 4 \le t \le 10? -

- - -

- It's helpful to think about the graph of y=r(t), which is given in Figure. -

-
- The rate r(t) of pollution leaking from a tank, measured in gallons per day. - -
-

- Since r(t) \ge 0, - the value of \int_4^{10} r(t) \, dt is the area under the curve on the interval [4,10]. - A Riemann sum for this area will have rectangles with heights measured in gallons per day and widths measured in days, - so the area of each rectangle will have units of - - \frac{\text{gallons} }{\text{day} } \cdot \text{days} = \text{gallons} - . -

- -

- Thus, the definite integral tells us the total number of gallons of pollutant that leak from the tank from day 4 to day 10. - The Total Change Theorem tells us the same thing: - if we let R(t) denote the total number of gallons of pollutant that have leaked from the tank up to day t, - then R'(t) = r(t), and - - \int_4^{10} r(t) \, dt = R(10) - R(4) - , - the number of gallons that have leaked from day 4 to day 10. -

- -

- To compute the exact value of the integral, - we use the Fundamental Theorem of Calculus. - Antidifferentiating r(t) = 0.0069t^3 -0.125t^2+11.079, - we find that - - \int_4^{10} 0.0069t^3 -0.125t^2+11.079 \, dt \amp= \left. 0.0069 \cdot \frac{1}{4} \, t^4 - 0.125 \cdot \frac{1}{3} t^3 + 11.079t \right|_4^{10} - \amp\approx 44.282 - . -

- -

- Thus, approximately 44.282 gallons of pollutant leaked over the six day time period. -

- -

- To find the average rate at which pollutant leaked from the tank over 4 \le t \le 10, - we compute the average value of r on [4,10]. - Thus, - - r_{\operatorname{AVG} [4,10]} = \frac{1}{10-4} \int_4^{10} r(t) \, dt \approx \frac{44.282}{6} = 7.380 - - gallons per day. -

- - - - - - - - Summary -

-

    -
  • -

    - We can find the exact value of a definite integral without taking the limit of a Riemann sum or using a familiar area formula by finding the antiderivative of the integrand, - and hence applying the Fundamental Theorem of Calculus. -

    -
    • - -
  • -

    - The Fundamental Theorem of Calculus says that if f is a continuous function on [a,b] and F is an antiderivative of f, then - - \int_a^b f(x) \, dx = F(b) - F(a) - . - Hence, - if we can find an antiderivative for the integrand f, - evaluating the definite integral comes from simply computing the change in F on [a,b]. -

    -
    • - -
  • -

    - A slightly different perspective on the FTC allows us to restate it as the Total Change Theorem, - which says that - - \int_a^b f'(x) \, dx = f(b) - f(a) - , - for any continuously differentiable function f. - This means that the definite integral of the instantaneous rate of change of a function f on an interval [a,b] is equal to the total change in the function f on [a,b]. -

    -
    • -
-

- - - + +