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The Second Fundamental Theorem of Calculus

In Section, we learned the Fundamental Theorem of Calculus (FTC), which from here forward will be referred to as the First Fundamental Theorem of Calculus, as in this section we develop a corresponding result that…

Introduction

In Section, we learned the Fundamental Theorem of Calculus (FTC), which from here forward will be referred to as the First Fundamental Theorem of Calculus, as in this section we develop a corresponding result that follows it. Recall that the First FTC tells us that if \(f\) is a continuous function on \([a,b]\) and \(F\) is any antiderivative of \(f\) (that is, \(F' = f\)), then \[\begin{aligned}\end{aligned}\].

We have used this result in two settings:

If we have a graph of \(f\) and we can compute the exact area bounded by \(f\) on an interval \([a,b]\), we can compute the change in an antiderivative \(F\) over the interval.
If we can find an algebraic formula for an antiderivative of \(f\), we can evaluate the integral to find the net signed area bounded by the function on the interval.

For the former, see Preview Activity or Activity. For the latter, we can easily evaluate exactly integrals such as \[\begin{aligned}\end{aligned}\], since we know that the function \(F(x) = \frac{1}{3}x^3\) is an antiderivative of \(f(x) = x^2\). Thus, \[\begin{aligned}\int_1^4 x^2 \, dx \amp= \frac{1}{3}x^3 \bigg\vert_1^4 \\ \amp= \frac{1}{3}(4)^3 - \frac{1}{3}(1)^3 \\ \amp= 21\end{aligned}\].

Thus, the First FTC can be used in two ways. First, to find the difference \(F(b) - F(a)\) for an antiderivative \(F\) of the integrand \(f\), even if we may not have a formula for \(F\) itself. To do this, we must know the value of the integral \(\int_a^b f(x) \, dx\) exactly, perhaps through known geometric formulas for area. In addition, the First FTC provides a way to find the exact value of a definite integral, and hence a certain net signed area exactly, by finding an antiderivative of the integrand and evaluating its total change over the interval. In this case, we need to know a formula for the antiderivative \(F\). Both of these perspectives are reflected in Figure.

The value of a definite integral may have additional meaning depending on context: as the change in position when the integrand is a velocity function, the total amount of pollutant leaked from a tank when the integrand is the rate at which pollution is leaking, or other total changes if the integrand is a rate function. Also, the value of the definite integral is connected to the average value of a continuous function on a given interval: \(f_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \int_a^b f(x) \, dx\).

Exploration

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

The Second Fundamental Theorem of Calculus

The result of Preview Activity is not particular to the function \(f(t) = 4-2t\), nor to the choice of \(1\) as the lower bound in the integral that defines the function \(A\). For instance, if we let \(f(t) = \cos(t) - t\) and set \(A(x) = \int_2^x f(t) \, dt\), we can determine a formula for \(A\) by the First FTC. Specifically, \[\begin{aligned}A(x) \amp= \int_2^x (\cos(t) - t) \, dt \\ \amp= \sin(t) - \frac{1}{2}t^2 \bigg\vert_2^x \\ \amp= \sin(x) - \frac{1}{2}x^2 - \left(\sin(2) - 2 \right)\end{aligned}\].

Differentiating \(A(x)\), since \((\sin(2) - 2)\) is constant, it follows that \[\begin{aligned}\end{aligned}\], and thus we see that \(A'(x) = f(x)\), so \(A\) is an antiderivative of \(f\). And since \(A(2) = \int_2^2 f(t) \, dt = 0\), \(A\) is the only antiderivative of \(f\) for which \(A(2) = 0\).

In general, if \(f\) is any continuous function, and we define the function \(A\) by the rule \[\begin{aligned}\end{aligned}\], where \(c\) is an arbitrary constant, then we can show that \(A\) is an antiderivative of \(f\). To see why, let's demonstrate that \(A'(x) = f(x)\) by using the limit definition of the derivative. Doing so, we observe that \[\begin{aligned}A'(x) \amp = \lim_{h \to 0} \frac{A(x+h) - A(x)}{h} \\ \amp = \lim_{h \to 0} \frac{\int_c^{x+h} f(t) \, dt - \int_c^x f(t) \, dt}{h} \\ \amp = \lim_{h \to 0} \frac{\int_x^{x+h} f(t) \, dt}{h}\end{aligned}\], where Equation follows from the fact that \(\int_c^x f(t) \,dt + \int_x^{x+h} f(t) \, dt = \int_c^{x+h} f(t) \, dt\). Now, observe that for small values of \(h\), \[\begin{aligned}\end{aligned}\], by a simple left-hand approximation of the integral. Thus, as we take the limit in Equation, it follows that \[\begin{aligned}\end{aligned}\].

Hence, \(A\) is indeed an antiderivative of \(f\). In addition, \(A(c) = \int_c^c f(t) \, dt = 0\). The preceding argument demonstrates the truth of the Second Fundamental Theorem of Calculus, which we state as follows.

If \(f\) is a continuous function and \(c\) is any constant, then \(f\) has a unique antiderivative \(A\) that satisfies \(A(c) = 0\), and that antiderivative is given by the rule \(A(x) = \int_c^x f(t) \, dt\).

Understanding Integral Functions

The Second FTC provides us with a way to construct an antiderivative of any continuous function. In particular, if we are given a continuous function \(g\) and wish to find an antiderivative \(G\), we can now say that \[\begin{aligned}\end{aligned}\] provides the rule for such an antiderivative, and moreover that \(G(c) = 0\). Note especially that we know that \(G'(x) = g(x)\), or \[\begin{aligned}\end{aligned}\].

This result is useful for understanding the graph of \(G\).

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

Differentiating an Integral Function

We have seen that the Second FTC enables us to construct an antiderivative \(F\) for any continuous function \(f\) as the integral function \(F(x) = \int_c^x f(t) \, dt\). If we have a function of the form \(F(x) = \int_c^x f(t) \, dt\), then we know that \(F'(x) = \frac{d}{dx} \left[\int_c^x f(t) \, dt \right] = f(x)\). This shows that integral functions, while perhaps having the most complicated formulas of any functions we have encountered, are nonetheless particularly simple to differentiate. For instance, if \[\begin{aligned}\end{aligned}\], then by the Second FTC, we know immediately that \[\begin{aligned}\end{aligned}\].

In general, we know by the Second FTC that \[\begin{aligned}\end{aligned}\].

This equation says that the derivative of the integral function whose integrand is \(f\), is \(f\). We see that if we first integrate the function \(f\) from \(t = a\) to \(t = x\), and then differentiate with respect to \(x\), these two processes undo each other.

What happens if we differentiate a function \(f(t)\) and then integrate the result from \(t = a\) to \(t = x\)? That is, what can we say about the quantity \[\begin{aligned}\end{aligned}\]

We note that \(f(t)\) is an antiderivative of \(\frac{d}{dt} \left[ f(t) \right]\) and apply the First FTC. We see that \[\begin{aligned}\int_a^x \frac{d}{dt} \left[ f(t) \right] \, dt \amp= f(t) \bigg\vert_a^x \\ \amp= f(x) - f(a)\end{aligned}\].

Thus, we see that if we first differentiate \(f\) and then integrate the result from \(a\) to \(x\), we return to the function \(f\), minus the constant value \(f(a)\). So the two processes almost undo each other, up to the constant \(f(a)\).

The observations made in the preceding two paragraphs demonstrate that differentiating and integrating (where we integrate from a constant up to a variable) are almost inverse processes. This should not be surprising: integrating involves antidifferentiating, which reverses the process of differentiating. On the other hand, we see that there is some subtlety involved, because integrating the derivative of a function does not quite produce the function itself. This is because every function has an entire family of antiderivatives, and any two of those antiderivatives differ only by a constant.

Summary

For a continuous function \(f\), the integral function \(A(x) = \int_1^x f(t) \, dt\) defines an antiderivative of \(f\).
The Second Fundamental Theorem of Calculus is the formal, more general statement of the preceding fact: if \(f\) is a continuous function and \(c\) is any constant, then \(A(x) = \int_c^x f(t) \, dt\) is the unique antiderivative of \(f\) that satisfies \(A(c) = 0\).
Together, the First and Second FTC enable us to formally see how differentiation and integration are almost inverse processes through the observations that \[\begin{aligned}\end{aligned}\] and \[\begin{aligned}\end{aligned}\].

Practice (4)

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

The tide removes sand from the beach at a small ocean park at a rate modeled by the function \[\begin{aligned}\end{aligned}\]
A pumping station adds sand to the beach at rate modeled by the function \[\begin{aligned}\end{aligned}\]
Both \(R(t)\) and \(S(t)\) are measured in cubic yards of sand per hour, \(t\) is measured in hours, and the valid times are \(0 \le t \le 6\). At time \(t = 0\), the beach holds 2500 cubic yards of sand.
1. What definite integral measures how much sand the tide will remove during the time period \(0 \le t \le 6\)? Why?
2. Write an expression for \(Y(x)\), the total number of cubic yards of sand on the beach at time \(x\). Carefully explain your thinking and reasoning.
3. At what instantaneous rate is the total number of cubic yards of sand on the beach at time \(t = 4\) changing?
4. Over the time interval \(0 \le t \le 6\), at approximately what time \(t\) is the amount of sand on the beach least? What is the corresponding approximate minimum value? Explain and justify your answers fully.
Odkrij odgovor
1. Since the tide removes sand at a rate of \(R(t) = 2 + 5\sin \left( \frac{4\pi t}{25} \right)\) cubic yards of sand per hour, on the time interval \([0,6]\), it follows that since the units on \(R(t) \cdot \triangle t\) are cubic yards, the total sand removed on this time interval is \[\begin{aligned}\end{aligned}\].
2. Because sand is being both removed and added, we can determine the total amount of sand that has been added or subtracted up to a given time \(t\) by integrating the difference between the the rates at which sand is being added and removed. In particular, since the total change in sand on the beach is the integral of the rate of change, we have that the total amount of sand on the beach at time \(x\) is given by the original amount (2500 cubic yards) plus the total change: \[\begin{aligned}\end{aligned}\].
3. We know that \(Y(x)\) measures the total amount of sand on the beach at time \(x\). Thus, the instantaneous rate at which the total number of cubic yards of sand on the beach is changing at time \(t=4\) is given by \(Y'(4)\). We first compute \(Y'(x)\). By the Second Fundamental Theorem of Calculus, we have \[\begin{aligned}\end{aligned}\], and thus \(Y'(4) = S(4) - R(4) \approx -1.90875\) cubic yards per hour.
4. To find where the total sand function \(Y(x)\) has its minimum value on \([0,6]\), we look at the behavior of \(Y'(x) = S(x) - R(x)\). Plotting \(Y'(x)\) on the interval, we see that \(Y'(x) = 0\) at \(x \approx 5.118\), and for \(0 \le x \lt 5.118\), \(Y'(x) \lt 0\), and for \(5.118 \lt x \le 6\), \(Y'(x) \gt 0\). Thus, \(Y\) changes from decreasing to increasing at \(x \approx 5.118\) and has an absolute minimum there on the interval \([0,6]\). Furthermore, \(Y(5.118) \approx 2492.368\).
Let \(g\) be the function pictured at left in Figure, and let \(F\) be defined by \(F(x) = \int_{2}^x g(t) \, dt\). Assume that the shaded areas have values \(A_1 = 4.29\), \(A_2 = 12.75\), \(A_3 = 0.36\), and \(A_4 = 1.79\). Assume further that the portion of \(A_2\) that lies between \(x = 0.5\) and \(x = 2\) is \(6.06\).
Sketch a carefully labeled graph of \(F\) on the axes provided, and include a written analysis of how you know where \(F\) is zero, increasing, decreasing, concave up, and concave down.

Odkrij odgovor

By the Second Fundamental Theorem of Calculus, we know that \(F'(x) = g(x)\). Thus, from the given graph of \(g\), it follows that \(F\) is increasing wherever \(g\) is positive (on the intervals \(x \lt -1\), \(0.5 \lt x \lt 4\), and \(5 \lt x \lt 6.5\)) and decreasing wherever \(g\) is negative (on \(-1 \lt x \lt 0.5\) and \(4 \lt x \lt 5\)). Furthermore, \(F\) is concave up wherever \(g\) is increasing (on approximately \(-0.4 \lt x \lt 2\) and \(4.5 \lt x \lt 6\)) and concave down werever \(g\) is decreasing (on approximately \(2 \lt x \lt 4.5\) and \(x \gt 6\)).
In addition, from the given data about definite integrals and properties of the definite integral, we can determine the following function values: \[\begin{aligned}F(2) \amp= \int_{2}^{2} g(t) \, dt = 0 \\ F(0.5) \amp= \int_{2}^{0.5} g(t) \, dt = -6.06 \\ F(4) \amp= \int_{2}^{4} g(t) \, dt = \int_{0.5}^4 g(t) \, dt - \int_{0.5}^2 g(t) \, dt = 12.75 - 6.06 = 6.69 \\ F(5) \amp= F(4) + \int_4^5 g(t) \,dt = 6.69 - 0.36 = 6.33 \\ F(6.5) \amp= F(5) + \int_4^{6.5} g(t) \,dt = 6.33 + 1.79 = 8.12 \\ F(-1) \amp= \int_2^{-1} g(t) \, dt = -\int_{-1}^2 g(t) \, dt = -\left( \int_{-1}^{0.5} g(t) \, dt + \int_{0.5}^2 g(t) \, dt \right) = -(-4.29 + 6.06) = -1.77\end{aligned}\]
In the figure below, we put all of this information to work and sketch an approximate graph of \(y = F(x)\).
The tide removes sand from the beach at a small ocean park at a rate modeled by the function \[\begin{aligned}\end{aligned}\]
A pumping station adds sand to the beach at rate modeled by the function \[\begin{aligned}\end{aligned}\]
Both \(R(t)\) and \(S(t)\) are measured in cubic yards of sand per hour, \(t\) is measured in hours, and the valid times are \(0 \le t \le 6\). At time \(t = 0\), the beach holds 2500 cubic yards of sand.
1. What definite integral measures how much sand the tide will remove during the time period \(0 \le t \le 6\)? Why?
2. Write an expression for \(Y(x)\), the total number of cubic yards of sand on the beach at time \(x\). Carefully explain your thinking and reasoning.
3. At what instantaneous rate is the total number of cubic yards of sand on the beach at time \(t = 4\) changing?
4. Over the time interval \(0 \le t \le 6\), at approximately what time \(t\) is the amount of sand on the beach least? What is the corresponding approximate minimum value? Explain and justify your answers fully.
Odkrij odgovor
1. Since the tide removes sand at a rate of \(R(t) = 2 + 5\sin \left( \frac{4\pi t}{25} \right)\) cubic yards of sand per hour, on the time interval \([0,6]\), it follows that since the units on \(R(t) \cdot \triangle t\) are cubic yards, the total sand removed on this time interval is \[\begin{aligned}\end{aligned}\].
2. Because sand is being both removed and added, we can determine the total amount of sand that has been added or subtracted up to a given time \(t\) by integrating the difference between the the rates at which sand is being added and removed. In particular, since the total change in sand on the beach is the integral of the rate of change, we have that the total amount of sand on the beach at time \(x\) is given by the original amount (2500 cubic yards) plus the total change: \[\begin{aligned}\end{aligned}\].
3. We know that \(Y(x)\) measures the total amount of sand on the beach at time \(x\). Thus, the instantaneous rate at which the total number of cubic yards of sand on the beach is changing at time \(t=4\) is given by \(Y'(4)\). We first compute \(Y'(x)\). By the Second Fundamental Theorem of Calculus, we have \[\begin{aligned}\end{aligned}\], and thus \(Y'(4) = S(4) - R(4) \approx -1.90875\) cubic yards per hour.
4. To find where the total sand function \(Y(x)\) has its minimum value on \([0,6]\), we look at the behavior of \(Y'(x) = S(x) - R(x)\). Plotting \(Y'(x)\) on the interval, we see that \(Y'(x) = 0\) at \(x \approx 5.118\), and for \(0 \le x \lt 5.118\), \(Y'(x) \lt 0\), and for \(5.118 \lt x \le 6\), \(Y'(x) \gt 0\). Thus, \(Y\) changes from decreasing to increasing at \(x \approx 5.118\) and has an absolute minimum there on the interval \([0,6]\). Furthermore, \(Y(5.118) \approx 2492.368\).
Let \(g\) be the function pictured at left in Figure, and let \(F\) be defined by \(F(x) = \int_{2}^x g(t) \, dt\). Assume that the shaded areas have values \(A_1 = 4.29\), \(A_2 = 12.75\), \(A_3 = 0.36\), and \(A_4 = 1.79\). Assume further that the portion of \(A_2\) that lies between \(x = 0.5\) and \(x = 2\) is \(6.06\).
Sketch a carefully labeled graph of \(F\) on the axes provided, and include a written analysis of how you know where \(F\) is zero, increasing, decreasing, concave up, and concave down.

Odkrij odgovor

By the Second Fundamental Theorem of Calculus, we know that \(F'(x) = g(x)\). Thus, from the given graph of \(g\), it follows that \(F\) is increasing wherever \(g\) is positive (on the intervals \(x \lt -1\), \(0.5 \lt x \lt 4\), and \(5 \lt x \lt 6.5\)) and decreasing wherever \(g\) is negative (on \(-1 \lt x \lt 0.5\) and \(4 \lt x \lt 5\)). Furthermore, \(F\) is concave up wherever \(g\) is increasing (on approximately \(-0.4 \lt x \lt 2\) and \(4.5 \lt x \lt 6\)) and concave down werever \(g\) is decreasing (on approximately \(2 \lt x \lt 4.5\) and \(x \gt 6\)).
In addition, from the given data about definite integrals and properties of the definite integral, we can determine the following function values: \[\begin{aligned}F(2) \amp= \int_{2}^{2} g(t) \, dt = 0 \\ F(0.5) \amp= \int_{2}^{0.5} g(t) \, dt = -6.06 \\ F(4) \amp= \int_{2}^{4} g(t) \, dt = \int_{0.5}^4 g(t) \, dt - \int_{0.5}^2 g(t) \, dt = 12.75 - 6.06 = 6.69 \\ F(5) \amp= F(4) + \int_4^5 g(t) \,dt = 6.69 - 0.36 = 6.33 \\ F(6.5) \amp= F(5) + \int_4^{6.5} g(t) \,dt = 6.33 + 1.79 = 8.12 \\ F(-1) \amp= \int_2^{-1} g(t) \, dt = -\int_{-1}^2 g(t) \, dt = -\left( \int_{-1}^{0.5} g(t) \, dt + \int_{0.5}^2 g(t) \, dt \right) = -(-4.29 + 6.06) = -1.77\end{aligned}\]
In the figure below, we put all of this information to work and sketch an approximate graph of \(y = F(x)\).

Symbols used here

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

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

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

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

\sin,\ \cos,\ \tan

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

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

derivative
Instantaneous rate of change; slope of the graph.

\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.

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: The Second Fundamental Theorem of Calculus

How does the integral function A(x) = \int_1^x f(t) \, dt define an antiderivative of f?
What is the statement of the Second Fundamental Theorem of Calculus?
How do the First and Second Fundamental Theorems of Calculus enable us to formally see how differentiation and integration are almost inverse processes?

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.

Več v Calculus

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