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Constructing accurate graphs of antiderivatives

A recurring theme in our discussion of differential calculus has been the question Given information about the derivative of an unknown function f, how much information can we obtain about f itself?

Introduction

A recurring theme in our discussion of differential calculus has been the question Given information about the derivative of an unknown function \(f\), how much information can we obtain about \(f\) itself? In Activity, the graph of \(y = f'(x)\) was known (along with the value of \(f\) at a single point) and we endeavored to sketch a possible graph of \(f\) near the known point. In Example, we investigated how the first derivative test enables us to use information about \(f'\) to determine where the original function \(f\) is increasing and decreasing, as well as where \(f\) has relative extreme values. If we know a formula or graph of \(f'\), by computing \(f''\) we can find where the original function \(f\) is concave up and concave down. Thus, knowing \(f'\) and \(f''\) enables us to understand the shape of the graph of \(f\).

We returned to this question in even more detail in Section. In that setting, we knew the instantaneous velocity of a moving object and worked to determine as much as possible about the object's position function. We found connections between the net signed area under the velocity function and the corresponding change in position of the function, and the Total Change Theorem further illuminated these connections between \(f'\) and \(f\), showing that the total change in the value of \(f\) over an interval \([a,b]\) is determined by the net signed area bounded by \(f'\) and the \(x\)-axis on the same interval.

In what follows, we explore the situation where we possess an accurate graph of the derivative function along with a single value of the function \(f\). From that information, we'd like to determine a graph of \(f\) that shows where \(f\) is increasing, decreasing, concave up, and concave down, and also provides an accurate function value at any point.

Exploration

Constructing the graph of an antiderivative

Preview Activity demonstrates that when we can find the exact area under the graph of a function on any given interval, it is possible to construct a graph of the function's antiderivative. That is, we can find a function whose derivative is given. We can now determine not only the overall shape of the antiderivative graph, but also the actual height of the graph at any point of interest.

This is a consequence of the Fundamental Theorem of Calculus: if we know a function \(f\) and the value of the antiderivative \(F\) at some starting point \(a\), we can determine the value of \(F(b)\) via the definite integral. Since \(F(b) - F(a) = \int_a^b f(x) \, dx\), it follows that \[\begin{aligned}\end{aligned}\].

We can also interpret the equation \(F(b) - F(a) = \int_a^b f(x) \, dx\) in terms of the graphs of \(f\) and \(F\) as follows. On an interval \([a,b]\),

differences in heights on the graph of the antiderivative given by \(F(b) - F(a)\) correspond to the net signed area bounded by the original function on the interval \([a,b]\), which is given by \(\int_a^b f(x) \, dx\).

Multiple antiderivatives of a single function

In the final question of Activity, we encountered a very important idea: a function \(f\) has more than one antiderivative. Each antiderivative of \(f\) is determined uniquely by its value at a single point. For example, suppose that \(f\) is the function given at left in Figure, and suppose further that \(F\) is an antiderivative of \(f\) that satisfies \(F(0) = 1\).

Then, using Equation, we can compute \[\begin{aligned}F(1) &= F(0) + \int_0^1 f(x) \, dx \\ &= 1 + 0.5 \\ &= 1.5\end{aligned}\]. Similarly, \(F(2) = 1.5\), \(F(3) = -0.5\), \(F(4) = -2\), \(F(5) = -0.5\), and \(F(6) = 1\). In addition, we can use the fact that \(F' = f\) to ascertain where \(F\) is increasing and decreasing, concave up and concave down, and has relative extremes and inflection points. We ultimately find that the graph of \(F\) is the one given in blue in Figure.

If we want an antiderivative \(G\) for which \(G(0) = 3\), then \(G\) will have the exact same shape as \(F\) (since both share the derivative \(f\)), but \(G\) will be shifted vertically from the graph of \(F\), as pictured in red in Figure. Note that \(G(1) - G(0) = \int_0^1 f(x) \, dx = 0.5\), just as \(F(1) - F(0) = 0.5\), but since \(G(0) = 3\), \(G(1) = G(0) + 0.5 = 3.5\), whereas \(F(1) = 1.5\). In the same way, if we assigned a different initial value to the antiderivative, say \(H(0) = -1\), we would get still another antiderivative, as shown in magenta in Figure.

This example demonstrates an important fact that holds more generally:

If \(G\) and \(H\) are both antiderivatives of a function \(f\) on an interval \((a,b)\), then the function \(G - H\) must be constant on the interval \((a,b)\).

To see why this result holds, observe that if \(G\) and \(H\) are both antiderivatives of \(f\), then \(G' = f\) and \(H' = f\). Hence, \[\begin{aligned}\end{aligned}\]. Since the only way a function can have derivative zero is by being a constant function, it follows that the function \(G - H\) must be constant.

We now see that if a function has at least one antiderivative, it must have infinitely many: we can add any constant of our choice to the antiderivative and get another antiderivative. For this reason, we sometimes refer to the general antiderivative of a function \(f\).

Condensed — the full section is in Boelkins, Active Calculus.

Functions defined by integrals

Equation allows us to compute the value of the antiderivative \(F\) at a point \(b\), provided that we know \(F(a)\) and can evaluate the definite integral from \(a\) to \(b\) of \(f\). That is, \[\begin{aligned}\end{aligned}\].

In several situations, we have used this formula to compute \(F(b)\) for several different values of \(b\), and then plotted the points \((b,F(b))\) to help us draw an accurate graph of \(F\). This suggests that we may want to think of \(b\), the upper limit of integration, as a variable itself. To that end, we introduce the idea of an integral function, a function whose formula involves a definite integral.

Note that because \(x\) is the independent variable in the function \(A\), and determines the endpoint of the interval of integration, we need to use a different variable as the variable of integration. A standard choice is \(t\), but any variable other than \(x\) is acceptable.

One way to think of the function \(A\) is as the net signed area from \(a\) up to \(x\) function, where we consider the region bounded by \(y = f(t)\). For example, in Figure, we see a function \(f\) pictured at left, and its corresponding area function (choosing \(a = 0\)), \(A(x) = \int_0^x f(t) \, dt\) shown at right.

The function \(A\) measures the net signed area from \(t = 0\) to \(t = x\) bounded by the curve \(y = f(t)\); this value is then reported as the corresponding height on the graph of \(y = A(x)\). This interactive graphic brings the static picture in Figure to life. There, the user can move the red point on the function \(f\) and see how the corresponding height changes at the light blue point on the graph of \(A\).

The choice of \(a\) is somewhat arbitrary. In the activity that follows, we explore how the value of \(a\) affects the graph of the integral function.

Summary

Given the graph of a function \(f\), we can construct the graph of its antiderivative \(F\) provided that (a) we know a starting value of \(F\), say \(F(a)\), and (b) we can evaluate the integral \(\int_a^b f(x) \, dx\) exactly for relevant choices of \(a\) and \(b\). For instance, if we wish to know \(F(3)\), we can compute \(F(3) = F(a) + \int_a^3 f(x) \, dx\). When we combine this information about the function values of \(F\) together with our understanding of how the behavior of \(F' = f\) affects the overall shape of \(F\), we can develop a completely accurate graph of the antiderivative \(F\).
Because the derivative of a constant is zero, if \(F\) is an antiderivative of \(f\), it follows that \(G(x) = F(x) + C\) will also be an antiderivative of \(f\). Moreover, any two antiderivatives of a function \(f\) differ precisely by a constant. Thus, any function with at least one antiderivative in fact has infinitely many, and the graphs of any two antiderivatives will differ only by a vertical translation.
Given a function \(f\), the rule \(A(x) = \int_a^x f(t) \, dt\) defines a new function \(A\) that measures the net-signed area bounded by \(f\) on the interval \([a,x]\). We call the function \(A\) the integral function corresponding to \(f\).

Practice (9)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Each of the functions on the left below could be described as having a specific algebraic structure as noted on the right. Match each function with its corresponding structure. It is possible that not all structures are used; and it is possible that not all functions have a description for its algebraic structure.
If you do not find a match in the previous exercise, explain why.
Describe the algebraic differences between the two functions \(f(x)=x^2 \sin x\) and \(g(x)=x^2- \sin x\).
A moving particle has its velocity given by the quadratic function \(v\) pictured in Figure. In addition, it is given that \(A_1 = \frac{7}{6}\) and \(A_2 = \frac{8}{3}\), as well as that for the corresponding position function \(s\), \(s(0) = 0.5\).
1. Use the given information to determine \(s(1)\), \(s(3)\), \(s(5)\), and \(s(6)\). What do these values mean in the context of the moving particle?
2. On what interval(s) is \(s\) increasing? That is, when is the particle moving forward? On what interval(s) is \(s\) decreasing? That is, when is the particle moving backward?
3. On what interval(s) is \(s\) concave up? That is, when is the particle accelerating? On what interval(s) is \(s\) concave down? That is, when is the particle decelerating?
4. Sketch an accurate, labeled graph of \(s\) on the axes at right in Figure.
5. Note that \(v(t) = -2 + \frac{1}{2}(t-3)^2\). Find a formula for \(s\).
Odkrij odgovor
1. We use the Fundamental Theorem of Calculus repeatedly together with the given information and the symmetry of the function \(v\), while paying careful attention to signed area:
  \(s(1) = s(0) + \int_0^1 v(t) \, dt = \frac{1}{2} + \frac{7}{6} = \frac{5}{3}\)
  \(s(3) = s(1) + \int_1^3 v(t) \, dt = \frac{5}{3} - \frac{8}{3} = -1\)
  \(s(5) = s(3) + \int_3^5 v(t) \, dt = -1 - \frac{8}{3} = -\frac{11}{3}\)
  \(s(6) = s(5) + \int_5^6 v(t) \, dt = -\frac{11}{3} + \frac{7}{6} = -\frac{5}{2}\)
2. The position function \(s\) is increasing on \(0 \lt t \lt 1\) and \(5 \lt t \lt 6\) since this is where \(v(t) = s'(t)\) is positive. Similarly, \(s\) is decreasing for \(1 \lt t \lt 5\).
3. The position function \(s\) is concave down for \(t \lt 3\) since \(v = s'\) is decreasing on this interval, and \(s\) is concave up for \(t \gt 3\) because \(v = s'\) increases there.
4. In the following figure, we see a plot of \(s\) that reflects all of the known information, including the points \((1, \frac{5}{3})\), \((3, -1)\), \((5, -\frac{11}{3})\), and \((6, -2.5)\).
5. The general antiderivative of \(v(t) = -2 + \frac{1}{2}(t-3)^2\) is \(s(t) = -2t + \frac{1}{6}(t-3)^3 + C\). Given that \(s(0) = \frac{1}{2}\), we know that \(\frac{1}{2} = \frac{1}{6}(-3)^3 + C\), and thus \(C = \frac{1}{2} + \frac{27}{6} = \frac{30}{6} = 5\). Hence, \(s(t) = -2t + \frac{1}{6}(t-3)^3 + 5\).
A person exercising on a treadmill experiences different levels of resistance and thus burns calories at different rates, depending on the treadmill's setting. In a particular workout, the rate at which a person is burning calories is given by the piecewise constant function \(c\) pictured in Figure. Note that the units on \(c\) are calories per minute.
1. Let \(C\) be an antiderivative of \(c\). What does the function \(C\) measure? What are its units?
2. Assume that \(C(0) = 0\). Determine the exact value of \(C(t)\) at the values \(t = 5, 10, 15, 20, 25, 30\).
3. Sketch an accurate graph of \(C\) on the axes provided at right in Figure. Be certain to label the scale on the vertical axis.
4. Determine a formula for \(C\) that does not involve an integral and is valid for \(5 \le t \le 10\).
Odkrij odgovor
1. An antiderivative function \(C\) measures the total number of calories burned in the workout since \(t = 0\). Since the units on \(c(t) = C'(t)\) are calories per minute, the units on \(C(t)\) are calories.
2. The value of \(C(5) = C(0) + \int_0^5 c(t) \, dt\). Since \(c\) is piecewise constant, the value of the integral is simply the area of the rectangle bounded by \(c\), so \(C(5) = C(0) + (2.5)(5) = 0 + 12.5 = 12.5\). Reasoning similarly,
  \(C(10) = C(5) + \int_{5}^{10} c(t) \, dt = 12.5 + (7.5)(5) = 50\)
  \(C(15) = C(10) + \int_{10}^{15} c(t) \, dt = 50 + (15)(5) = 125\)
  \(C(20) = C(15) + \int_{15}^{20} c(t) \, dt = 125 + (12.5)(5) = 187.5\)
  \(C(25) = C(20) + \int_{20}^{25} c(t) \, dt = 187.5 + (10)(5) = 237.5\)
  \(C(30) = C(25) + \int_{25}^{30} c(t) \, dt = 237.5 + (5)(5) = 262.5\)
3. In the figure below, we see that \(C\) is a piecewise linear function that passes through the points we established in (b) (such as \((15, 125)\) and \((20, 187.5)\)) and that the slope of \(C\) on each interval where \(c\) is constant is simply the value of \(c(t)\) on that interval (such as \(m = 12.5\) on \(15 \lt t \lt 20\)).
4. \(C\) is linear on \(5 \le t \le 10\) with slope \(m = 7.5\) and passes through the point \((5, C(5)) = (5, 12.5)\). Thus, \(C(t) = 12.5 + 7.5(t-5)\) on this interval.

Consider the piecewise linear function \(f\) given in Figure. Let the functions \(A\), \(B\), and \(C\) be defined by the rules \(A(x) = \int_{-1}^{x} f(t) \, dt\), \(B(x) = \int_{0}^{x} f(t) \, dt\), and \(C(x) = \int_{1}^{x} f(t) \, dt\).

For the values \(x = -1, 0, 1, \ldots, 6\), make a table that lists corresponding values of \(A(x)\), \(B(x)\), and \(C(x)\).
On the axes provided in Figure, sketch the graphs of \(A\), \(B\), and \(C\).
How are the graphs of \(A\), \(B\), and \(C\) related?
How would you best describe the relationship between the function \(A\) and the function \(f\)?

Odkrij odgovor

We use the First Fundamental Theorem and areas under the curves to complete the desired table. As an example, first note that \[\begin{aligned}\end{aligned}\]. Then, since the net-signed area between the graph of \(f\) and the \(x\)-axis on the interval \([-1,0]\) is the area of a square with side lengths of \(1\), we have \[\begin{aligned}\end{aligned}\] and thus \(B(0) - B(-1) = 1\), so \(0 - B(-1) = 1\) and \(B(-1) = -1\). \[\begin{aligned}B(1) &= B(0) + \int_0^1 f(t) \, dt = 0 + \frac{1}{2} = \frac{1}{2}, \\ B(2) &= B(1) + \int_1^2 f(t) \, dt = \frac{1}{2} + \left(-\frac{1}{2}\right) = 0, \\ B(3) &= B(2) + \int_2^3 f(t) \, dt = 0 + (-1) = -1, \\ B(4) &= B(3) + \int_3^4 f(t) \, dt = -1 + \left(-\frac{1}{2}\right) = -\frac{3}{2}, \\ B(5) &= B(4) + \int_4^5 f(t) \, dt = -\frac{3}{2} + \frac{1}{2} = -1, \\ B(6) &= B(5) + \int_5^6 f(t) \, dt = -1 + 1 = 0\end{aligned}\]. We can also say that \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\]. This allows us to complete the following table.

\(x\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)
\(A(x)\)	\(0\)	\(1\)	\(1.5\)	\(1\)	\(0\)	\(-0.5\)	\(0\)	\(1\)
\(B(x)\)	\(-1\)	\(0\)	\(0.5\)	\(0\)	\(-1\)	\(-1.5\)	\(-1\)	\(0\)
\(C(x)\)	\(-1.5\)	\(-0.5\)	\(0\)	\(-0.5\)	\(-1.5\)	\(-2\)	\(-1.5\)	\(-0.5\)

The graphs are shown the figure below. Note that \(A\), \(B\), and \(C\) are all differentiable functions (that are antiderivatives of \(f\), so each of their derivatives is \(f\)), so the graphs of these functions should be (and are) smooth.
Notice that the graphs of \(A\), \(B\), and \(C\) are vertical translations of each other. More specifically, \(A(x) = B(x) + 1\) and \(B(x) = C(x) + 0.5\).
Since \(A\) is an antiderivative of \(f\), we can say that \(A' = f\).

A moving particle has its velocity given by the quadratic function \(v\) pictured in Figure. In addition, it is given that \(A_1 = \frac{7}{6}\) and \(A_2 = \frac{8}{3}\), as well as that for the corresponding position function \(s\), \(s(0) = 0.5\).
1. Use the given information to determine \(s(1)\), \(s(3)\), \(s(5)\), and \(s(6)\). What do these values mean in the context of the moving particle?
2. On what interval(s) is \(s\) increasing? That is, when is the particle moving forward? On what interval(s) is \(s\) decreasing? That is, when is the particle moving backward?
3. On what interval(s) is \(s\) concave up? That is, when is the particle accelerating? On what interval(s) is \(s\) concave down? That is, when is the particle decelerating?
4. Sketch an accurate, labeled graph of \(s\) on the axes at right in Figure.
5. Note that \(v(t) = -2 + \frac{1}{2}(t-3)^2\). Find a formula for \(s\).
Odkrij odgovor
1. We use the Fundamental Theorem of Calculus repeatedly together with the given information and the symmetry of the function \(v\), while paying careful attention to signed area:
  \(s(1) = s(0) + \int_0^1 v(t) \, dt = \frac{1}{2} + \frac{7}{6} = \frac{5}{3}\)
  \(s(3) = s(1) + \int_1^3 v(t) \, dt = \frac{5}{3} - \frac{8}{3} = -1\)
  \(s(5) = s(3) + \int_3^5 v(t) \, dt = -1 - \frac{8}{3} = -\frac{11}{3}\)
  \(s(6) = s(5) + \int_5^6 v(t) \, dt = -\frac{11}{3} + \frac{7}{6} = -\frac{5}{2}\)
2. The position function \(s\) is increasing on \(0 \lt t \lt 1\) and \(5 \lt t \lt 6\) since this is where \(v(t) = s'(t)\) is positive. Similarly, \(s\) is decreasing for \(1 \lt t \lt 5\).
3. The position function \(s\) is concave down for \(t \lt 3\) since \(v = s'\) is decreasing on this interval, and \(s\) is concave up for \(t \gt 3\) because \(v = s'\) increases there.
4. In the following figure, we see a plot of \(s\) that reflects all of the known information, including the points \((1, \frac{5}{3})\), \((3, -1)\), \((5, -\frac{11}{3})\), and \((6, -2.5)\).
5. The general antiderivative of \(v(t) = -2 + \frac{1}{2}(t-3)^2\) is \(s(t) = -2t + \frac{1}{6}(t-3)^3 + C\). Given that \(s(0) = \frac{1}{2}\), we know that \(\frac{1}{2} = \frac{1}{6}(-3)^3 + C\), and thus \(C = \frac{1}{2} + \frac{27}{6} = \frac{30}{6} = 5\). Hence, \(s(t) = -2t + \frac{1}{6}(t-3)^3 + 5\).
A person exercising on a treadmill experiences different levels of resistance and thus burns calories at different rates, depending on the treadmill's setting. In a particular workout, the rate at which a person is burning calories is given by the piecewise constant function \(c\) pictured in Figure. Note that the units on \(c\) are calories per minute.
1. Let \(C\) be an antiderivative of \(c\). What does the function \(C\) measure? What are its units?
2. Assume that \(C(0) = 0\). Determine the exact value of \(C(t)\) at the values \(t = 5, 10, 15, 20, 25, 30\).
3. Sketch an accurate graph of \(C\) on the axes provided at right in Figure. Be certain to label the scale on the vertical axis.
4. Determine a formula for \(C\) that does not involve an integral and is valid for \(5 \le t \le 10\).
Odkrij odgovor
1. An antiderivative function \(C\) measures the total number of calories burned in the workout since \(t = 0\). Since the units on \(c(t) = C'(t)\) are calories per minute, the units on \(C(t)\) are calories.
2. The value of \(C(5) = C(0) + \int_0^5 c(t) \, dt\). Since \(c\) is piecewise constant, the value of the integral is simply the area of the rectangle bounded by \(c\), so \(C(5) = C(0) + (2.5)(5) = 0 + 12.5 = 12.5\). Reasoning similarly,
  \(C(10) = C(5) + \int_{5}^{10} c(t) \, dt = 12.5 + (7.5)(5) = 50\)
  \(C(15) = C(10) + \int_{10}^{15} c(t) \, dt = 50 + (15)(5) = 125\)
  \(C(20) = C(15) + \int_{15}^{20} c(t) \, dt = 125 + (12.5)(5) = 187.5\)
  \(C(25) = C(20) + \int_{20}^{25} c(t) \, dt = 187.5 + (10)(5) = 237.5\)
  \(C(30) = C(25) + \int_{25}^{30} c(t) \, dt = 237.5 + (5)(5) = 262.5\)
3. In the figure below, we see that \(C\) is a piecewise linear function that passes through the points we established in (b) (such as \((15, 125)\) and \((20, 187.5)\)) and that the slope of \(C\) on each interval where \(c\) is constant is simply the value of \(c(t)\) on that interval (such as \(m = 12.5\) on \(15 \lt t \lt 20\)).
4. \(C\) is linear on \(5 \le t \le 10\) with slope \(m = 7.5\) and passes through the point \((5, C(5)) = (5, 12.5)\). Thus, \(C(t) = 12.5 + 7.5(t-5)\) on this interval.

For the values \(x = -1, 0, 1, \ldots, 6\), make a table that lists corresponding values of \(A(x)\), \(B(x)\), and \(C(x)\).
On the axes provided in Figure, sketch the graphs of \(A\), \(B\), and \(C\).
How are the graphs of \(A\), \(B\), and \(C\) related?
How would you best describe the relationship between the function \(A\) and the function \(f\)?

Odkrij odgovor

\(x\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)
\(A(x)\)	\(0\)	\(1\)	\(1.5\)	\(1\)	\(0\)	\(-0.5\)	\(0\)	\(1\)
\(B(x)\)	\(-1\)	\(0\)	\(0.5\)	\(0\)	\(-1\)	\(-1.5\)	\(-1\)	\(0\)
\(C(x)\)	\(-1.5\)	\(-0.5\)	\(0\)	\(-0.5\)	\(-1.5\)	\(-2\)	\(-1.5\)	\(-0.5\)

The graphs are shown the figure below. Note that \(A\), \(B\), and \(C\) are all differentiable functions (that are antiderivatives of \(f\), so each of their derivatives is \(f\)), so the graphs of these functions should be (and are) smooth.
Notice that the graphs of \(A\), \(B\), and \(C\) are vertical translations of each other. More specifically, \(A(x) = B(x) + 1\) and \(B(x) = C(x) + 0.5\).
Since \(A\) is an antiderivative of \(f\), we can say that \(A' = f\).

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Constructing accurate graphs of antiderivatives

Given the graph of a function's derivative, how can we construct a completely accurate graph of the original function?
How many antiderivatives does a given function have? What do those antiderivatives all have in common?
Given a function f, how does the rule A(x) = \int_0^x f(t) \, dt define a new function A?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Poskusi sam.

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests