# 5.2: The Second Fundamental Theorem of Calculus

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##### Motivating Questions
• How does the integral function $$A(x) = \int_1^x f(t) \, dt$$ define an antiderivative of $$f\text{?}$$
• 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?

In Section 4.4, 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

$\int_a^b f(x) \, dx = F(b) - F(a)\text{.} \nonumber$

We have used this result in two settings:

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

For the former, see Preview Activity 5.1.1 or Activity 5.1.2. For the latter, we can easily evaluate exactly integrals such as

$\int_1^4 x^2 \, dx\text{,} \nonumber$

since we know that the function $$F(x) = \frac{1}{3}x^3$$ is an antiderivative of $$f(x) = x^2\text{.}$$ Thus,

\begin{align*} \int_1^4 x^2 \, dx &= \frac{1}{3}x^3 \bigg\vert_1^4\4pt] &= \frac{1}{3}(4)^3 - \frac{1}{3}(1)^3\\[4pt] &= 21\text{.} \end{align*} Thus, the First FTC can used in two ways. First, to find the difference $$F(b) - F(a)$$ for an antiderivative $$F$$ of the integrand $$f\text{,}$$ 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 $$F\text{.}$$ Both of these perspectives are reflected in Figure $$\PageIndex{1}$$.  Figure $$\PageIndex{1}$$. At left, the graph of $$f(x) = x^2$$ on the interval $$[1,4]$$ and the area it bounds. At right, the antiderivative function $$F(x) = \frac{1}{3}x^3\text{,}$$ whose total change on $$[1,4]$$ is the value of the definite integral at left. 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: $$f_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \int_a^b f(x) \, dx\text{.}$$ In the last part of Section 5.1, we studied integral functions of the form $$A(x) = \int_c^x f(t) \, dt\text{.}$$ Figure 5.1.5 is a particularly important image to keep in mind as we work with integral functions, and the corresponding java applet at gvsu.edu/s/cz can help us understand the function $$A\text{.}$$ In what follows, we use the First FTC to gain additional understanding of the function $$A(x) = \int_c^x f(t) \, dt\text{,}$$ where the integrand $$f$$ is given (either through a graph or a formula), and $$c$$ is a constant. ##### Preview Activity $$\PageIndex{1}$$ Consider the function $$A$$ defined by the rule \[ A(x) = \int_1^x f(t) \, dt\text{,} \nonumber

where $$f(t) = 4-2t\text{.}$$

1. Compute $$A(1)$$ and $$A(2)$$ exactly.
2. Use the First Fundamental Theorem of Calculus to find a formula for $$A(x)$$ that does not involve integrals. That is, use the first FTC to evaluate $$\int_1^x (4-2t) \, dt\text{.}$$
3. Observe that $$f$$ is a linear function; what kind of function is $$A\text{?}$$
4. Using the formula you found in (b) that does not involve integrals, compute $$A'(x)\text{.}$$
5. While we have defined $$f$$ by the rule $$f(t) = 4-2t\text{,}$$ it is equivalent to say that $$f$$ is given by the rule $$f(x) = 4 - 2x\text{.}$$ What do you observe about the relationship between $$A$$ and $$f\text{?}$$

## The Second Fundamental Theorem of Calculus

The result of Preview Activity $$\PageIndex{1}$$1 is not particular to the function $$f(t) = 4-2t\text{,}$$ nor to the choice of “$$1$$” as the lower bound in the integral that defines the function $$A\text{.}$$ For instance, if we let $$f(t) = \cos(t) - t$$ and set $$A(x) = \int_2^x f(t) \, dt\text{,}$$ we can determine a formula for $$A$$ by the First FTC. Specifically,

\begin{align*} A(x) &= \int_2^x (\cos(t) - t) \, dt\4pt] &= \sin(t) - \frac{1}{2}t^2 \bigg\vert_2^x\\[4pt] &= \sin(x) - \frac{1}{2}x^2 - \left(\sin(2) - 2 \right)\text{.} \end{align*} Differentiating $$A(x)\text{,}$$ since $$(\sin(2) - 2)$$ is constant, it follows that \[ A'(x) = \cos(x) - x\text{,} \nonumber

and thus we see that $$A'(x) = f(x)\text{,}$$ so $$A$$ is an antiderivative of $$f\text{.}$$ And since $$A(2) = \int_2^2 f(t) \, dt = 0\text{,}$$ $$A$$ is the only antiderivative of $$f$$ for which $$A(2) = 0\text{.}$$

In general, if $$f$$ is any continuous function, and we define the function $$A$$ by the rule

$A(x) = \int_c^x f(t) \, dt\text{,} \nonumber$

where $$c$$ is an arbitrary constant, then we can show that $$A$$ is an antiderivative of $$f\text{.}$$ 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{align} A'(x) & = \lim_{h \to 0} \frac{A(x+h) - A(x)}{h}\notag\4pt] & = \lim_{h \to 0} \frac{\int_c^{x+h} f(t) \, dt - \int_c^x f(t) \, dt}{h}\notag\\[4pt] & = \lim_{h \to 0} \frac{\int_x^{x+h} f(t) \, dt}{h}\text{,}\label{E-FTC2limdef}\tag{$$\PageIndex{1}$$} \end{align} where Equation ($$\PageIndex{1}$$) follows from the fact that $$\int_c^x f(t) \,dt + \int_x^{x+h} f(t) \, dt = \int_c^{x+h} f(t) \, dt\text{.}$$ Now, observe that for small values of $$h\text{,}$$ \[ \int_x^{x+h} f(t) \, dt \approx f(x) \cdot h\text{,} \nonumber

by a simple left-hand approximation of the integral. Thus, as we take the limit in Equation ($$\PageIndex{1}$$), it follows that

$A'(x) = \lim_{h \to 0} \frac{\int_x^{x+h} f(t) \, dt}{h} = \lim_{h \to 0} \frac{f(x) \cdot h}{h} = f(x)\text{.} \nonumber$

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

##### The Second Fundamental Theorem of Calculus

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

##### Activity $$\PageIndex{2}$$

Suppose that $$f$$ is the function given in Figure $$\PageIndex{2}$$ and that $$f$$ is a piecewise function whose parts are either portions of lines or portions of circles, as pictured.  Figure $$\PageIndex{2}$$. At left, the graph of $$y = f(x)\text{.}$$ At right, axes for sketching $$y = A(x)\text{.}$$

In addition, let $$A$$ be the function defined by the rule $$A(x) = \int_2^x f(t) \, dt\text{.}$$

1. What does the Second FTC tell us about the relationship between $$A$$ and $$f\text{?}$$
2. Compute $$A(1)$$ and $$A(3)$$ exactly.
3. Sketch a precise graph of $$y = A(x)$$ on the axes at right that accurately reflects where $$A$$ is increasing and decreasing, where $$A$$ is concave up and concave down, and the exact values of $$A$$ at $$x = 0, 1, \ldots, 7\text{.}$$
4. How is $$A$$ similar to, but different from, the function $$F$$ that you found in Activity 5.1.2?
5. With as little additional work as possible, sketch precise graphs of the functions $$B(x) = \int_3^x f(t) \, dt$$ and $$C(x) = \int_1^x f(t) \, dt\text{.}$$ Justify your results with at least one sentence of explanation.

## 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 $$G\text{,}$$ we can now say that

$G(x) = \int_c^x g(t) \, dt \nonumber$

provides the rule for such an antiderivative, and moreover that $$G(c) = 0\text{.}$$ Note especially that we know that $$G'(x) = g(x)\text{,}$$ or

$\frac{d}{dx} \left[ \int_c^x g(t) \, dt \right] = g(x)\text{.}\label{aTe}\tag{$$\PageIndex{2}$$}$

This result is useful for understanding the graph of $$G\text{.}$$

##### Example $$\PageIndex{1}$$

Investigate the behavior of the integral function

$E(x) = \int_0^x e^{-t^2} \, dt\text{.} \nonumber$

$$E$$ is closely related to the well known error function  1  in probability and statistics. It turns out that the function $$e^{-t^2}$$ does not have an elementary antiderivative.

The error function is defined by the rule $$\erf (x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \,dt$$ and has the key property that $$0 \le \erf (x) \lt 1$$ for all $$x \ge 0$$ and moreover that $$\lim_{x \to \infty} \erf (x) = 1\text{.}$$

While we cannot evaluate $$E$$ exactly for any value other than $$x = 0\text{,}$$ we still can gain a tremendous amount of information about the function $$E\text{.}$$ By applying the rule in Equation ($$\PageIndex{2}$$) to $$E\text{,}$$ it follows that

$E'(x) = \frac{d}{dx} \left[ \int_0^x e^{-t^2} \, dt \right] = e^{-x^2}\text{,} \nonumber$

so we know a formula for the derivative of $$E\text{,}$$ and we know that $$E(0) = 0\text{.}$$ This information is precisely the type we were given in Activity 3.1.2, where we were given information about the derivative of a function, but lacked a formula for the function itself.

Using the first and second derivatives of $$E\text{,}$$ along with the fact that $$E(0) = 0\text{,}$$ we can determine more information about the behavior of $$E\text{.}$$ First, we note that for all real numbers $$x\text{,}$$ $$e^{-x^2} \gt 0\text{,}$$ and thus $$E'(x) \gt 0$$ for all $$x\text{.}$$ Thus $$E$$ is an always increasing function. Further, as $$x \to \infty\text{,}$$ $$E'(x) = e^{-x^2} \to 0\text{,}$$ so the slope of the function $$E$$ tends to zero as $$x \to \infty$$ (and similarly as $$x \to -\infty$$). Indeed, it turns out that $$E$$ has horizontal asymptotes as $$x$$ increases or decreases without bound.

In addition, we can observe that $$E''(x) = -2xe^{-x^2}\text{,}$$ and that $$E''(0) = 0\text{,}$$ while $$E''(x) \lt 0$$ for $$x \gt 0$$ and $$E''(x) \gt 0$$ for $$x \lt 0\text{.}$$ This information tells us that $$E$$ is concave up for $$x\lt 0$$ and concave down for $$x \gt 0$$ with a point of inflection at $$x = 0\text{.}$$

The only thing we lack at this point is a sense of how big $$E$$ can get as $$x$$ increases. If we use a midpoint Riemann sum with 10 subintervals to estimate $$E(2)\text{,}$$ we see that $$E(2) \approx 0.8822\text{;}$$ a similar calculation to estimate $$E(3)$$ shows little change ($$E(3) \approx 0.8862$$), so it appears that as $$x$$ increases without bound, $$E$$ approaches a value just larger than $$0.886\text{,}$$ which aligns with the fact that $$E$$ has horizontal asymptotes. Putting all of this information together (and using the symmetry of $$f(t) = e^{-t^2}$$), we see the results shown in Figure $$\PageIndex{3}$$. Because $$E$$ is the antiderivative of $$f(t) = e^{-t^2}$$ that satisfies $$E(0) = 0\text{,}$$ values on the graph of $$y = E(x)$$ represent the net signed area of the region bounded by $$f(t) = e^{-t^2}$$ from 0 up to $$x\text{.}$$ We see that the value of $$E$$ increases rapidly near zero but then levels off as $$x$$ increases, since there is less and less additional accumulated area bounded by $$f(t) = e^{-t^2}$$ as $$x$$ increases.

##### Activity $$\PageIndex{3}$$

Suppose that $$f(t) = \frac{t}{1+t^2}$$ and $$F(x) = \int_0^x f(t) \, dt\text{.}$$

1. On the axes at left in Figure $$\PageIndex{4}$$, plot a graph of $$f(t) = \frac{t}{1+t^2}$$ on the interval $$-10 \le t \le 10\text{.}$$ Clearly label the vertical axes with appropriate scale.
2. What is the key relationship between $$F$$ and $$f\text{,}$$ according to the Second FTC?
3. Use the first derivative test to determine the intervals on which $$F$$ is increasing and decreasing.
4. Use the second derivative test to determine the intervals on which $$F$$ is concave up and concave down. Note that $$f'(t)$$ can be simplified to be written in the form $$f'(t) = \frac{1-t^2}{(1+t^2)^2}\text{.}$$
5. Using technology appropriately, estimate the values of $$F(5)$$ and $$F(10)$$ through appropriate Riemann sums.
6. Sketch an accurate graph of $$y = F(x)$$ on the righthand axes provided, and clearly label the vertical axes with appropriate scale.  Figure $$\PageIndex{4}$$. Axes for plotting $$f$$ and $$F\text{.}$$

## 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 $$F(x) = \int_c^x f(t) \, dt\text{.}$$ If we have a function of the form $$F(x) = \int_c^x f(t) \, dt\text{,}$$ then we know that $$F'(x) = \frac{d}{dx} \left[\int_c^x f(t) \, dt \right] = f(x)\text{.}$$ 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

$F(x) = \int_{\pi}^x \sin(t^2) \, dt\text{,} \nonumber$

then by the Second FTC, we know immediately that

$F'(x) = \sin(x^2)\text{.} \nonumber$

In general, we know by the Second FTC that

$\frac{d}{dx} \left[ \int_a^x f(t) \, dt \right] = f(x)\text{.} \nonumber$

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

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

$\int_a^x \frac{d}{dt} \left[ f(t) \right] \, dt? \nonumber$

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{align*} \int_a^x \frac{d}{dt} \left[ f(t) \right] \, dt &= f(t) \bigg\vert_a^x\4pt] &= f(x) - f(a)\text{.} \end{align*} Thus, we see that if we first differentiate $$f$$ and then integrate the result from $$a$$ to $$x\text{,}$$ we return to the function $$f\text{,}$$ minus the constant value $$f(a)\text{.}$$ So the two processes almost undo each other, up to the constant $$f(a)\text{.}$$ 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. ##### Activity $$\PageIndex{4}$$ Evaluate each of the following derivatives and definite integrals. Clearly cite whether you use the First or Second FTC in so doing. 1. $$\displaystyle \frac{d}{dx} \left[ \int_4^x e^{t^2} \, dt \right]$$ 2. $$\displaystyle \int_{-2}^x \frac{d}{dt} \left[ \frac{t^4}{1+t^4} \right] \, dt$$ 3. $$\displaystyle \frac{d}{dx} \left[ \int_{x}^1 \cos(t^3) \, dt \right]$$ 4. $$\displaystyle \int_{3}^x \frac{d}{dt} \left[ \ln(1+t^2) \right] \, dt$$ 5. $$\frac{d}{dx} \left[ \int_4^{x^3} \sin(t^2) \, dt \right]$$. ## Summary • For a continuous function $$f\text{,}$$ the integral function $$A(x) = \int_1^x f(t) \, dt$$ defines an antiderivative of $$f\text{.}$$ • 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 $$A(c) = 0\text{.}$$ • Together, the First and Second FTC enable us to formally see how differentiation and integration are almost inverse processes through the observations that \[ \int_c^x \frac{d}{dt} \left[ f(t) \right] \, dt = f(x) - f(c) \nonumber

and

$\frac{d}{dx} \left[ \int_c^x f(t) \, dt \right] = f(x)\text{.} \nonumber$

This page titled 5.2: The Second Fundamental Theorem of Calculus is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Matthew Boelkins, David Austin & Steven Schlicker (ScholarWorks @Grand Valley State University) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.