2.4: The Fundamental Theorem of Integrals

The main theorem of this section is key to understanding the importance of definite integrals. In particular, we will invoke it in developing new applications for definite integrals. Moreover, we will use it to verify the fundamental theorem of calculus.

We first need some new notation and terminology. Suppose \(\epsilon\) is a nonzero infinitesimal. Intuitively, \(\epsilon\) is infinitely smaller than any nonzero real number. One way to express this is to note that for any nonzero real number \(r\), \[\frac{\epsilon}{r} \simeq 0 ,\] that is, the ratio of \(\epsilon\) to \(r\) is an infinitesimal. Now we also have \[\frac{\epsilon^{2}}{\epsilon}=\epsilon \simeq 0 ,\] that is, the ratio of \(\epsilon^{2}\) to \(\epsilon\) is an infinitesimal. Intuitively, this means that \(\epsilon^{2}\) is infinitely smaller than \(\epsilon\) itself. This is related to a fact about real numbers: For any real number \(r\) with \(0<r<1, r^{2}\) is smaller than \(r .\) For example, if \(r=0.01,\) then \(r^{2}=0.0001\).

Definition

Given a nonzero hyperreal number \(\epsilon,\) we say another hyperreal number \(\delta\) is of an order less than \(\epsilon\) if \(\frac{\delta}{\epsilon}\) is an infinitesimal, in which case we write \(\delta \sim o(\epsilon) .\)

In other words, we have \[\delta \sim o(\epsilon) \text { if and only if } \frac{\delta}{\epsilon} \simeq 0 .\]

Example \(\PageIndex{1}\)

If \(\alpha\) is any infinitesimal, than \(\alpha \sim o(1)\) since \(\frac{\alpha}{1}=\alpha\) is an infinitesimal.

Example \(\PageIndex{2}\)

If \(\epsilon\) is any nonzero infinitesimal, then \(\epsilon^{2} \sim o(\epsilon)\) since

\[\frac{\epsilon^{2}}{\epsilon}=\epsilon \simeq 0 .\]

Now suppose \(N\) is a positive infinite integer, \(\epsilon=\frac{1}{N},\) and \(\delta_{i} \sim o(\epsilon)\) for \(i=1,2, \ldots, N .\) Then, for any positive real number \(r,\)

\[\frac{\left|\delta_{i}\right|}{\epsilon}<r ,\] and so \[\sum_{i=1}^{N} \frac{\left|\delta_{i}\right|}{\epsilon}<r N .\] Multiplying both sides by \(\epsilon,\) we have \[\sum_{i=1}^{N}\left|\delta_{i}\right|<r N \epsilon=r N \frac{1}{N}=r .\] Since this holds for all positive real numbers \(r,\) it follows that \(\sum_{i=1}^{N}\left|\delta_{i}\right|\) is an infinitesimal. Now \[\left|\sum_{i=1}^{N} \delta_{i}\right| \leq \sum_{i=1}^{N}\left|\delta_{i}\right| ,\] and so we may conclude that \(\sum_{i=1}^{n} \delta_{i}\) is an infinitesimal. In words, the sum of \(N\) infinitesimals of order less than \(\frac{1}{N}\) is still an infinitesimal. Now suppose \(B\) is a function that for any real numbers \(a<b\) in an open interval \(I\) assigns a value \(B(a, b) .\) Moreover, suppose \(B\) has the following two properties: • for any \(a<c<b\) in \(I, B(a, b)=B(a, c)+B(c, b),\) and • for some continuous function \(h\) and any nonzero infinitesimal \(d x\), \[B(x, x+d x)-h(x) d x \sim o(d x) \] for any \(x\) in \(I\). For a positive infinite integer \(N,\) let \(d x=\frac{b-a}{N}\) and let \[a=x_{0}<x_{1}<x_{2}<\cdots<x_{N}=b \] be a partition of \([a, b]\) into \(N\) equal intervals of length \(d x .\) Then \[\begin{aligned} B(a, b) &=B\left(x_{0}, x_{1}\right)+B\left(x_{1}, x_{2}\right)+B\left(x_{2}, x_{3}\right)+\cdots+B\left(x_{N-1}, x_{N}\right) \\ &=\sum_{i=1}^{N} B\left(x_{i-1}, x_{i-1}+d x\right) \\ &\left.=\sum_{i=1}^{N}\left(B\left(x_{i-1}, x_{i-1}+d x\right)-h\left(x_{i-1}\right) d x\right)+h\left(x_{i-1}\right) d x\right) \\ &=\sum_{i=1}^{N}\left(B\left(x_{i-1}, x_{i-1}+d x\right)-h\left(x_{i-1}\right) d x\right)+\sum_{i=1}^{N} h\left(x_{i-1}\right) d x \\ & \simeq \sum_{i=1}^{N}\left(B\left(x_{i-1}, x_{i-1}+d x\right)-h\left(x_{i-1}\right) d x\right)+\int_{a}^{b} h(x) d x .\end{aligned} \] Since the final sum on the right is the sum of \(N\) infinitesimals of order less than \(\frac{1}{N},\) it follows that \[B(a, b)=\int_{a}^{b} h(x) d x .\] This result is basic to understanding both the computation of definite integrals and their applications. We call it the fundamental theorem of integrals.

Theorem \(\PageIndex{1}\)

Suppose \(B\) is a function that for any real numbers \(a<b\) in an open interval \(I\) assigns a value \(B(a, b)\) and satisfies

• for any \(a<c<b\) in \(I, B(a, b)=B(a, c)+B(c, b),\) and • for some continuous function \(h\) and any nonzero infinitesimal \(d x\), \[B(x, x+d x)-h(x) d x \sim o(d x)\] for any \(x\) in \(I\). Then \[B(a, b)=\int_{a}^{b} h(x) d x\] for any real numbers \(a\) and \(b\) in \(I\).

We will look at several applications of definite integrals in the next section. For now, we note how this theorem provides a method for evaluating integrals. Namely, given a function \(f\) which is differentiable on an open interval \(I,\) define, for every \(a<b\) in \(I\),

\[B(a, b)=f(b)-f(a).\] Then, for any \(a, b,\) and \(c\) in \(I\) with \(a<c<b\), \[\begin{aligned} B(a, b) &=f(b)-f(a) \\ &=(f(b)-f(c))+(f(c)-f(a)) \\ &=B(a, c)+B(c, b). \end{aligned}\] Moreover, for any infinitesimal \(d x\) and any \(x\) in \(I\), \[\frac{B(x, x+d x)}{d x}=\frac{f(x+d x)-f(x)}{d x} \simeq f^{\prime}(x),\] from which it follows that \[\frac{B(x, x+d x)-f^{\prime}(x) d x}{d x}\] is an infinitesimal. Hence \[B(x, d x)-f^{\prime}(x) d x \sim o(d x),\] and so it follows from Theorem 2.4.1 that \[f(b)-f(a)=B(a, b)=\int_{a}^{b} f^{\prime}(x) d x,\] This is the fundamental theorem of calculus.

Theorem \(\PageIndex{2}\)

If \(f\) is differentiable on an open interval \(I,\) then for every \(a<b\) in \(I\),

\[\int_{a}^{b} f^{\prime}(x) d x=f(b)-f(a).\]

Example \(\PageIndex{3}\)

To evaluate

\[\int_{0}^{1} x d x,\] we first note that \(g(x)=x\) is the derivative of \(f(x)=\frac{1}{2} x^{2} .\) Hence, by Theorem \(2.4 .2,\) \[\int_{0}^{1} x d x=f(1)-f(0)=\frac{1}{2}-0=\frac{1}{2}\] We will write \[\left.f(x)\right|_{a} ^{b}=f(b)-f(a)\] to simplify the notation for evaluating an integral using Theorem \(2.4 .2 .\) With this notation, the previous example becomes \[\int_{0}^{1} x d x=\left.\frac{1}{2} x^{2}\right|_{0} ^{1}=\frac{1}{2}-0=\frac{1}{2}.\]

Example \(\PageIndex{4}\)

Since

\[\int x^{2} d x=\frac{1}{3} x^{3}+c,\] we have \[\int_{1}^{2} x^{2} d x=\left.\frac{1}{3} x^{3}\right|_{1} ^{2}=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}.\]

Example \(\PageIndex{5}\)

Since

\[\int-20 \sin (5 x) d x=4 \cos (5 x)+c,\] we have \[\int_{0}^{2 \pi}-20 \sin (5 t) d t=\left.4 \cos (5 t)\right|_{0} ^{2 \pi}=4-4=0.\] Note that if we consider an object moving along a straight line with velocity \(v(t)=-20 \sin (5 t),\) then this definite integral computes the change in position of the object from time \(t=0\) to time \(t=2 \pi .\) In this case, the object, although always in motion, is in the same position at time \(t=2 \pi\) as it was at time \(t=2 \pi\).

Exercise \(\PageIndex{1}\)

Evaluate \(\int_{0}^{1} x^{4} d x\).

Answer

\(\int_{0}^{1} x^{4} d x=\frac{1}{5}\)

Exercise \(\PageIndex{2}\)

Evaluate \(\int_{0}^{\pi} \sin (x) d x\).

Answer

\(\int_{0}^{\pi} \sin (x) d x=2\)

Exercise \(\PageIndex{3}\)

Suppose the velocity of an object moving along a straight line is \(v(t)=10 \sin (t)\) centimeters per second. Find the change in position of the object from time \(t=0\) to time \(t=\pi\).

Answer

20 centimeters