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Section8.6The Fundamental Theorem of Calculus

When we introduced the definite integral, we also learned about accumulation functions. An accumulation function is a function \(A\) defined as a definite integral from a fixed lower limit \(a\) to a variable upper limit where the integrand is a given rate function \(f\text{,}\)

\begin{equation*} A(x) = \int_a^x f(z)\, dz. \end{equation*}

The motivation for such a function was that the definite integral computes the total change of a quantity when the rate of change is given.

The Fundamental Theorem of Calculus provides our guarantee that this makes valid mathematical sense. In particular, we will learn that the derivative of an accumulation function is the same rate function that was used to define the accumulation. Then knowing this fact will allow us to compute any definite integral for which we know an antiderivative of the integrand.

The proof relies on the Mean Value Theorem for Integrals. Given the accumulation function \(A(x)\) and its associated integrand \(f(x)\text{,}\) we focus on the definition of the derivative for \(A(x)\text{:}\)

\begin{equation*} A'(x) = \lim_{h \to 0} \frac{A(x+h)-A(x)}{h}. \end{equation*}

We use the splitting property of definite integrals to rewrite this as

\begin{equation*} A'(x) = \lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(z) \, dz. \end{equation*}

Because the function inside the limit is an average value of \(f\) on the interval from \(x\) to \(x+h\) and \(f\) is continuous, the mean value theorem allows us to replace that formula with the value of \(f\) at some point \(c\) between \(x\) and \(x+h\text{,}\) and we treat \(c\) as a function of \(h\text{:}\)

\begin{equation*} A'(x) = \lim_{h \to 0} f(c(h)). \end{equation*}

As \(h \to 0\text{,}\) since \(c\) is constrained to be between \(x\) and \(x+h\text{,}\) we know \(c \to x\) and continuity allows us to conclude

\begin{equation*} A'(x) = \lim_{h \to 0} f(c(h)) = f(x). \end{equation*}

Most calculus textbooks test understanding of this concept with problems relating to differentiation, including the use of the chain rule.

Example8.6.2

Compute the following derivatives.

  1. \(\displaystyle \frac{d}{dx} \int_1^x e^{-z^2} \, dz\)
  2. \(\displaystyle \frac{d}{dx} \int_0^{\sqrt{x}} \frac{1}{\sqrt{z^4+1}} \, dz\)
  3. \(\displaystyle \frac{d}{dx} \int_{x}^{x^2} \sin(\cos(z))\, dz\)
Solution

When solving problems involving definite integrals, it is often helpful to explicitly remind yourself of the concept of accumulation functions and the fundamental theorem of calculus's conclusion.

  1. Define \(\displaystyle A(x) = \int_1^x e^{-z^2} \, dz\text{,}\) which is the accumulation function with integrand \(f(z) = e^{-z^2}\text{.}\) The Fundamental Theorem of Calculus tells us that the accumulation function is an antiderivative of the integrand, which means \(A'(x) = f(x) = e^{-x^2}\text{,}\) where we evaluate with the variable \(x\) instead of the variable of integration. The following work communicates this.

    \begin{align*} A(x) &= \int_1^x e^{-z^2} \, dz\\ A'(x) &\overset{{\mathrm{FTC}}}{=} e^{-x^2}\\ \frac{d}{dx} \int_1^x e^{-z^2} \, dz &= A'(x) = e^{-x^2} \end{align*}
  2. To compute \(\displaystyle \frac{d}{dx}\left[ \int_0^{\sqrt{x}} \frac{1}{\sqrt{z^4+1}}\,dz \right]\text{,}\) we will need to define an accumulation function and then apply the Fundamental Theorem of Calculus to find the derivative required. The integral given is not the accumulation function, but we use the same integrand with an upper limit that is just the input variable.

    \begin{align*} A(x) &= \int_0^x \frac{1}{\sqrt{z^4+1}} \, dz\\ A'(x)&\overset{{\mathrm{FTC}}}{=} \frac{1}{\sqrt{x^4+1}}\\ \frac{d}{dx}\left[ \int_0^{\sqrt{x}} \frac{1}{\sqrt{z^4+1}}\,dz \right] &= \frac{d}{dx}[A(\sqrt{x})]\\ &= A'(\sqrt{x}) \frac{d}{dx}[\sqrt{x}] \\ &= \frac{1}{\sqrt{(\sqrt{x})^4+1}} \cdot (\frac{1}{2}x^{-1/2}) \\ &= \frac{1}{\sqrt{x^2+1}} \cdot \frac{1}{2 \sqrt{x}} \\ &= \frac{1}{2 \sqrt{x(x^2+1)}} \end{align*}
  3. The accumulation function requires a constant in the lower limit, although that constant doesn't matter. Define \(\displaystyle A(x) = \int_1^x \sin(\cos(z))\,dz\text{,}\) although the choice of \(a=1\) was arbitrary. We will rewrite the formula in question using the splitting property of definite integrals and then apply the chain rule.

    \begin{align*} A(x) &= \int_1^x \sin(\cos(z)) \, dz\\ A'(x) & \overset{{\mathrm{FTC}}}{=} \sin(\cos(x)) \\ \frac{d}{dx} \int_x^{x^2} \sin(\cos(z)) \, dz &= \frac{d}{dx}\left[\int_1^{x^2} \sin(\cos(z))\,dz - \int_1^{x} \sin(\cos(z))\,dz \right]\\ &= \frac{d}{dx}[A(x^2) - A(x)] \\ &= A'(x^2) \cdot \frac{d}{dx}[x^2] - A'(x) \\ &= \sin(\cos(x^2)) \cdot (2x) - \sin(\cos(x))\\ &= 2x \sin(\cos(x^2)) - \sin(\cos(x)) \end{align*}

Conceptually, it is more important to recognize that definite integrals represent evaluation of accumulation functions and know that those accumulation functions are antiderivatives of the integrand. This leads to the second half of the Fundamental Theorem of Calculus, the evaluation of definite integrals using any known antiderivative.

This theorem just puts three key concepts to work with one another. First, we define an accumulation function. In order to deal with the possibility that \(a\) or \(b\) might be dependent variables and not constants, let \(c\) be a constant value in \(I\text{.}\) We use this value as the lower limit of an accumulation function,

\begin{equation*} A(x) = \int_c^{x} f(z)\, dz, \end{equation*}

defined for any value \(x \in I\text{.}\)

The Fundamental Theorem of Calculus guarantees \(A(x)\) is an antiderivative of \(f(x)\text{.}\) The hypothesis is that \(F(x)\) is also an antiderivative of \(f(x)\text{,}\) defined on the same interval \(I\text{.}\) The second key concept is that any two antiderivatives defined on the same interval can only differ by a constant. That is, there is a real number \(K\) so that \(F(x) = A(x)+K\text{.}\)

Third, we use the splitting property of definite integrals, which says

\begin{equation*} \int_a^b f(x) \, dx = \int_c^b f(x)\, dx - \int_c^a f(x) \, dx. \end{equation*}

These two integrals are equivalent to the evaluation of our accumulation function, so that we can rewrite

\begin{equation*} \int_a^b f(x) \, dx = A(b) - A(a). \end{equation*}

However, any two antiderivatives defined on the same interval will change by the same amount,

\begin{equation*} F(b)-F(a) = \big(A(b)+K\big) - \big(A(a)+K\big) = A(b)-A(a). \end{equation*}

This proves our result

\begin{equation*} \int_a^b f(x)\,dx = F(b) - F(a). \end{equation*}

The evaluation of definite integrals using the change in an antiderivative occurs so frequently that a notation has been adopted to represent this process. Given any function \(F(x)\text{,}\) the notation \(\big[ F(x) \big]_{a}^{b}\) means evaluate the change of the expression when \(x\) goes from \(a\) to \(b\text{:}\)

\begin{equation*} \big[ F(x) \big]_{a}^{b} = F(b) - F(a). \end{equation*}

When the formula for \(F(x)\) is simple, the brackets can be dropped and replaced by a vertical bar on the right,

\begin{equation*} F(x) \big|_a^b = F(b) - F(a). \end{equation*}

Applications of the Fundamental Theorem of Calculus to evaluate a definite integral generally involve two steps. First, identify an antiderivative. Second, evaluate the change in that antiderivative. The evaluation notation described above allows us to represent these two steps when we acknowledge we are using this theorem.

Example8.6.4

Evaluate \(\displaystyle \int_1^4 x^2 \, dx\text{.}\)

Solution

Let us consider the two steps separately. Then we will see how to represent this more compactly using evaluation notation.

First, we need an antiderivative, computed as an indefinite integral

\begin{equation*} \int x^2 \, dx = \frac{1}{3}x^3 + C. \end{equation*}

This tells us that \(F(x) = \frac{1}{3}x^3\) is an antiderivative, as is \(F(x) = \frac{1}{3}x^3+4\) (or any other constant). We'll use the first, since it is simpler; we only need one antiderivative, not all of them.

Second, the Fundamental Theorem of Calculus allows us to evaluate the definite integral as the change in \(F(x)\text{.}\)

\begin{equation*} \int_1^4 x^2 \, dx = F(4)-F(1) = \frac{1}{3}(4)^3 - \frac{1}{3}(1)^3 = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21 \end{equation*}

This can be written more compactly by writing the formula of the antiderivative inside the evaluation notation and indicating that we are using the Fundamental Theorem of Calculus by annotating the equal sign by “FTC”:

\begin{align*} \int_{1}^{4} x^2 \,dx &\overset{{\mathrm{FTC}}}{=} \big[ \frac{1}{3}x^3 \big]_1^4 \\ &= \frac{1}{3}(4)^3 - \frac{1}{3}(1)^3 = \frac{64}{3} - \frac{1}{3} = 21 \end{align*}

Notice that we did not need the constant of integration because the Fundamental Theorem of Calculus only requires one antiderivative, and we generally wish to choose the most convenient one that does not have a constant term.

Evaluation of definite integrals involves recognizing antiderivatives and then evaluating their change.

Example8.6.5

\begin{align*} \displaystyle \int_0^\pi \sin(x) \, dx & \overset{{\mathrm{FTC}}}{=} \big[-\cos(x)\big]_{0}^{\pi}\\ &= -\cos(\pi) - -\cos(0) \\ &= -(-1) - -(1) = 2 \end{align*}
Example8.6.6

\begin{align*} \displaystyle \int_1^2 e^{3x} \, dx & \overset{{\mathrm{FTC}}}{=} \big[\frac{1}{3}e^{3x}\big]_{1}^{2}\\ &= \frac{1}{3}e^{6} - \frac{1}{3}e^{3} \\ &= \frac{e^6-e^3}{3} \end{align*}