2.2: Definite Integrals

We will now consider another approach to solving the problem of finding the position function given a velocity function. As above, suppose \(v(t)\) specifies, at time \(t,\) the velocity of an object moving along a straight line, starting at time \(t=a\) and ending at time \(t=b .\) Let \(x(t)\) be the position of the object at time \(t\) and let \(x_{0}=x(a)\) be the initial position of the object.

Recall that if an object travels at constant velocity \(r\) for a time \(T,\) then its change in position is simply \(r T\) (to the right if \(r>0\) and to the left if \(r<0\) ). It follows that if \(v(t)\) were constant, say \(v(t)=r\) for some fixed real number \(r\) and all \(t\) in \([a, b],\) then

\[x(t)=x(a)+r(t-a)=x_{0}+r(t-a) \]

since the object will have been traveling, after starting at \(x_{0},\) at a velocity \(r\) for a time period of length \(t-a .\)

If \(v(t)\) is not constant, but doesn't vary by much, then \((2.2 .1)\) will give a good approximation to \(x(t) .\) In general, \(v\) may change significantly over the interval, but, as long as \(v\) is a continuous function, we can subdivide \([a, b]\) into small intervals for which \(v(t)\) does not change by much over any given subinterval. That is, if we choose points \(t_{0}, t_{1}, t_{2}, \ldots, t_{n}\) such that

\[a=t_{0}<t_{1}<t_{2}<\cdots<t_{n}=b ,\]

let \(\Delta t_{i}=t_{i}-t_{i-1}\) for \(i=1,2,3, \ldots, n,\) and choose real numbers \(t_{1}^{*}, t_{2}^{*}, t_{3}^{*}, \ldots,\) \(t_{n}^{*}\) so that \(t_{i-1} \leq t_{i}^{*} \leq t_{i},\) then \(v\left(t_{i}^{*}\right) \Delta t_{i} \) will approximate well the change of position of the object from time \(t=t_{i-1}\) to time \(t=t_{i},\) provided \(v\) is continuous and \(\Delta t_{i}\) is small. It follows that we may approximate the position of the object at time \(t=b\) by adding together all the approximate changes in position over the subintervals. That is,

\[x(b) \approx x(a)+v\left(t_{1}^{*}\right) \Delta t_{1}+v\left(t_{2}^{*}\right) \Delta t_{2}+\cdots+v\left(t_{n}^{*}\right) \Delta t_{n} .\]

Example \(\PageIndex{1}\)

In an earlier example, we had \(v(t)=-20 \sin (5 t)\) centimeters per second and \(x(0)=4\) centimeters. To approximate \(x(2),\) we will divide \([0,2]\) into four equal subintervals, each of length \(0.5 .\) That is, we will take

\[t_{0}=0.0, t_{1}=0.5, t_{2}=1, t_{3}=1.5, t_{4}=2,\]

and

\[\Delta t_{1}=0.5, \Delta t_{2}=0.5, \Delta t_{3}=0.5, \Delta t_{4}=0.5.\]

Good choices for points to evaluate \(v(t)\) are the midpoints of the subintervals. In this case, that means we should take

\[t_{1}^{*}=0.25, t_{2}^{*}=0.75, t_{3}^{*}=1.25, t_{4}^{*}=1.75.\]

Then we have

\[\begin{aligned} x(2) &\approx x(0)+v(0.25) \Delta t_{1}+v(0.75) \Delta t_{2}+v(1.25) \Delta t_{3}+v(1.75) \Delta t_{4} \\ &= 4-20 \sin (1.25)(0.5)-20 \sin (3.75)(0.5)-20 \sin (6.25)(0.5) \\ &-20 \sin (8.75)(0.5) \\ &\approx-5.6897, \end{aligned}\]

Note that, from our earlier work, we know that the exact answer is

\[x(2)=4 \cos (10) \approx-3.3563.\]

We may improve upon our approximation by using smaller subintervals. For example, if we divide \([0,2]\) into 10 equal subintervals, each of length \(0.2,\) then we would have

\[\begin{array}{l}{t_{0}=0.0, t_{1}=0.2, t_{2}=0.4, t_{3}=0.6, t_{4}=0.8, t_{5}=1.0,} \\ {t_{6}=1.2, t_{7}=1.4, t_{8}=1.6, t_{9}=1.8, t_{10}=2.0}\end{array}\]

and

\[\Delta t_{1}=\Delta t_{2}=\cdots=\Delta t_{10}=0.2.\]

If we evaluate \(v(t)\) at the midpoints again, then we take

\[\begin{array}{l}{t_{1}^{*}=0.1, t_{2}^{*}=0.3, t_{3}^{*}=0.5, t_{4}^{*}=0.7, t_{5}^{*}=0.9,} \\ {t_{6}^{*}=1.1, t_{7}^{*}=1.3, t_{8}^{*}=1.5, t_{9}^{*}=1.7, t_{10}^{*}=1.9.}\end{array}\]

Hence we have

\[\begin{aligned} x(2) & \approx x(0)+v(0.1) \Delta t_{1}+v(0.3) \Delta t_{2}+v(0.5) \Delta t_{3}+\cdots+v(1.9) \Delta t_{10} \\ &=4+0.2(v(0.1)+v(0.3)+v(0.5)+\cdots+v(1.9)) \\ &=4+0.2(-20 \sin (0.5)-20 \sin (1.5)-20 \sin (2.5)-\cdots-20 \sin (9.5)) \\ &\approx-3.6720, \end{aligned}\]

a significant improvement over our first approximation.

Exercise \(\PageIndex{1}\)

Suppose the velocity of an object at time \(t\) is \(v(t)=10 \sin (t)\) centimeters per second. Let \(x(t)\) be the position of the object at time \(t .\) If \(x(0)=10\) centimeters, use the technique of the previous example to approximate \(x(3)\) using \((\mathrm{a}) n=6\) and \((\mathrm{b}) n=12\) subintervals.

Answer

(a) 30.11 centimeters, (b) 29.95 centimeters

One question which arises immediately is why we would want to find approximations to the position function when we already know how to find the position function exactly using an integral. There are two answers. First, it is not always possible to find an integral for a given function, even when an integral does exist. For example, if the velocity function in the previous example were \(v(t)=-20 \sin \left(5 t^{2}\right),\) then it would not be possible to write an expression for the integral of \(v\) in terms of the elementary functions of calculus. In this case, the best we could do is look for good approximations for the position function \(x .\)

Second, this approach also leads to an exact expression for the position function. If we let \(N\) be an infinitely large positive integer and divide \([a, b]\) into an infinite number of equal subintervals of infinitesimal length

\[d t=\frac{b-a}{N},\]

then we should expect that

\[x(b) \simeq x(a)+v\left(t_{1}^{*}\right) d t+v\left(t_{2}^{*}\right) d t+\cdots+v\left(t_{N}^{*}\right) d t,\]

where, similar to the work above, \(t_{i}^{*}\) is a hyperreal number in the \(i\) th subinterval. Rewriting this as

\[x(b)-x(a) \simeq v\left(t_{1}^{*}\right) d t+v\left(t_{2}^{*}\right) d t+\cdots+v\left(t_{N}^{*}\right) d t,\]

we are saying that the change in position of the object from time \(t=a\) to time \(t=b\) is equal to an infinite sum of infinitesimal changes. Although Zeno was correct in saying that an infinite sum of zeros is still zero, an infinite sum of infinitesimal values need not be infinitesimal.

The right-hand side of \((2.2 .7)\) provides the motivation for the following definition.

Definition

Suppose \(f\) is a continuous function on a closed bounded interval \([a, b] .\) Given a positive integer \(N,\) finite or infinite, we call a set of numbers \(\left\{t_{0}, t_{1}, t_{2}, \cdots, t_{n}\right\}\) a partition of \([a, b]\) if

\[a=t_{0}<t_{1}<t_{2}<\cdots<t_{N}=b.\]

Given such a partition of \([a, b],\) let \(t_{i}^{*}\) denote a number with \(t_{i-1} \leq t_{i}^{*} \leq t_{i}\) and

let \(\Delta t_{i}=t_{i}-t_{i-1},\) where \(i=1,2,3, \ldots, N .\) We call a sum of the form

\[\sum_{i=1}^{N} f\left(t_{i}^{*}\right) \Delta t_{i}=f\left(t_{1}^{*}\right) \Delta t_{1}+f\left(t_{2}^{*}\right) \Delta t_{2}+f\left(t_{3}^{*}\right) \Delta t_{3}+\cdots+f\left(t_{N}^{*}\right) \Delta t_{N}\]

a Riemann sum. If \(N\) is infinite and the partition forms subintervals of equal length

\[d t=\Delta t_{1}=\Delta t_{2}=\cdots=\Delta t_{n},\]

then we call the shadow of the Riemann sum the definite integral of \(f\) from \(a\) to \(b,\) which we denote

\[\int_{a}^{b} f(t) d t.\]

That is,

\[\int_{a}^{b} f(t) d t=\operatorname{sh}\left(\sum_{i=1}^{N} f\left(t_{i}^{*}\right) d t\right).\]

Note that in order for \((2.2 .12)\) to make sense, we need the Riemann sum on the right-hand side to be finite (so it has a shadow) and to have the same shadow for all choices of infinite integers \(N\) and evaluation points \(t_{i}^{*}\) (so the definite integral has a unique value). These are both true for continuous functions on closed bounded intervals, but the verifications would take us into subtle properties of continuous functions beyond the scope of this text.

In terms of velocity and position functions, if \(v(t)\) is the velocity and \(x(t)\) the position of an object moving along a straight line from time \(t=a\) to time \(t=b, \text { we may now write (combining }(2.2 .7) \text { with }(2.2 .12))\)

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

Since \(v\) is the derivative of \(x,\) we might ask if \((2.2 .13)\) is true for any differentiable function. That is, if \(f\) is differentiable on an interval which contains the real numbers \(a\) and \(b,\) is it always the case that

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

This is in fact true, but requires more explanation than just the intuitive approach of our example with positions and velocities. We will come back to this important result, which we call the fundamental theorem of calculus, after developing some properties of definite integrals.