Trapezoidal Rule

The \(N\)th trapezoidal approximation to \(\displaystyle\int_a^b f(x)\, dx\) is \[ \boxed{\bbox[#FAF8ED,5pt]{ T_N = \frac{1}{2}\,\Delta x \bigl(y_0 + 2y_1 + \cdots + 2y_{N-1} + y_N \bigr) }} \]

where \(\Delta x = \dfrac{b-a}{N}\) and \(y_j = f(a+j\,\Delta x)\).

CONCEPTUAL INSIGHT

We see in Figure 7.25 that the area of the \(j\)th trapezoid is equal to the average of the areas of the endpoint rectangles with heights \(y_{j-1}\) and \(y_j\). It follows that \(T_N\) is equal to the average of the right- and left-endpoint approximations \(R_N\) and \(L_N\) introduced in Section 5.1: \[ T_N=\frac12(R_N+L_N) \]

In general, this average is a better approximation than either \(R_N\) alone or \(L_N\) alone.

EXAMPLE 1

Calculate \(T_8\) for the integral \({ \int_1^{3} \sin(x^2)\, dx}\). Then use a computer algebra system to calculate \(T_N\) for \(N = 50, 100, 500, 1000\), and \(10{,}000\).

Solution Divide \([1,3]\) into \(N=8\) subintervals of length \(\Delta x = \frac{3-1}{8} = \frac{1}{4}\). Then sum the function values at the endpoints (Figure 7.26) with the appropriate coefficients: \[ \begin{eqnarray*} T_{8}&=& \frac12\left(\frac14\right)\big[\sin(1^2) + 2\sin(1.25^2) + 2\sin(1.5^2) + 2\sin(1.75^2)\\ &&\quad{}+ 2\sin(2^2) + 2\sin(2.25^2) + 2\sin(2.5^2) + 2\sin(2.75^2) + \sin(3^2)\big]\\ &\approx& 0.4281 \end{eqnarray*} \]

Figure 7.26: Division on \([1,3]\) into \(N=8\) subintervals.

In general, \(\Delta x = (3-1)/N = 2/N\) and \(x_j = 1+ 2j/N\). In summation notation, \[ T_N = \frac12\left(\frac2N\right)\Biggl[\sin(1^2)+2 \underbrace{\sum_{j=1}^{N-1}\sin\left(\left(1+\frac{2j}N\, \right)^2\right)}_{\textrm{Sum of terms with coefficient \(2\)}}+\sin(3^2) \Biggr] \]

We evaluate the inner sum on a CAS, using a command such as
\(\mathtt{Sum[Sin[(1+2j/N)}\textrm{^}\)\(\mathtt{2], \{j, 1,N-1\}]}\)

The results in Table 7.1 suggest that \({\int_1^3 \sin (x^2)\,dx}\) is approximately 0.4633.

Midpoint Rule

The \(N\)th midpoint approximation to \(\displaystyle\int_a^b f(x)\, dx\) is \[ \boxed{\bbox[#FAF8ED,5pt]{ M_N = \Delta x \bigl( f(c_1) + f(c_2) +\cdots + f(c_N) \bigr)}} \]

where \(\Delta x = \dfrac{b-a}{N}\) and \(c_j = a + \big(j-\frac12\big)\,\Delta x\) is the midpoint of \([x_{j-1},x_j]\).

GRAPHICAL INSIGHT

\(M_N\) has a second interpretation as the sum of the areas of tangential trapezoids—that is, trapezoids whose top edges are tangent to the graph of \(f(x)\) at the midpoints \(c_j\) [Figure 7.28]. The trapezoids have the same area as the rectangles because the top edge of the trapezoid passes through the midpoint of the top edge of the rectangle, as shown in Figure 7.27.

In the error bound, you can let \(K_2\) be the maximum of \(|f''(x)|\) on \([a,b]\), but if it is inconvenient to find this maximum exactly, take \(K_2\) to be any number that is definitely larger than the maximum.

THEOREM 1 Error Bound for \(T_N\) and \(M_N\)

Assume \(f''(x)\) exists and is continuous. Let \(K_2\) be a number such that \(|f''(x)| \le K_2\) for all \(x\in[a,b]\). Then \[ \boxed{\bbox[#FAF8ED,5pt]{\text{Error}(T_N) \le \frac{K_2(b- a)^3}{12N^2},\qquad \text{Error}(M_N) \le \frac{K_2(b- a)^3}{24N^2}}} \]

GRAPHICAL INSIGHT

Note that the error bound for \(M_N\) is one-half of the error bound for \(T_N\), suggesting that \(M_N\) is generally more accurate than \(T_N\). Why do both error bounds depend on \(f''(x)\)? The second derivative measures concavity, so if \(|f''(x)|\) is large, then the graph of \(f\) bends a lot and trapezoids do a poor job of approximating the region under the graph. Thus the errors in both \(T_N\) and \(M_N\) (which uses tangential trapezoids) are likely to be large (Figure 7.29).

EXAMPLE 2 Checking the Error Bound

Calculate \(T_6\) and \(M_6\) for \(\displaystyle\int_1^4 \sqrt{x}\, dx\).

1. Calculate the error bounds.
2. Calculate the integral exactly and verify that the error bounds are satisfied.

Solution Divide \([1,4]\) into six subintervals of width \(\Delta x =\frac{4-1}{6} = \frac12\). Using the endpoints and midpoints shown in Figure 7.30, we obtain \[ \begin{align*} T_6&= \frac12\left(\frac12\right)\Big(\sqrt{1} + 2\sqrt{1.5} + 2\sqrt{2} + 2\sqrt{2.5} + 2\sqrt{3} + 2\sqrt{3.5} + \sqrt{4} \Big)\approx 4.661488 \\ M_6&= \frac12 \Big(\sqrt{1.25} + \sqrt{1.75} + \sqrt{2.25} + \sqrt{2.75} + \sqrt{3.25} + \sqrt{3.75}\Big) \approx 4.669245 \end{align*} \]

Figure 7.30: Interval \([1,4]\) divided into \(N=6\) subintervals.

1. Let \(f(x) = \sqrt x\). We must find a number \(K_2\) such that \(|f''(x)|\le K_2\) for \(1\le x \le 4\). We have \(f''(x)=-\frac14x^{-3/2}\). The absolute value \(|f''(x)|=\frac14x^{-3/2}\) is decreasing on \([1,4]\), so its maximum occurs at \(x=1\) (Figure 7.31). Thus, we may take \(K_2=|f''(1)|=\frac14\). By Theorem 1, \[ \begin{align*} \text{Error}(T_6) &\le \frac{K_2(b- a)^3}{12N^2} = \frac{\frac14(4-1)^3}{12(6)^2} =\frac1{64}\approx 0.0156 \\ \text{Error}(M_6) &\le \frac{K_2(b- a)^3}{24N^2} = \frac{\frac14(4 - 1)^3}{24(6)^2} = \frac1{128} \approx 0.0078 \end{align*} \]
   

   Figure 7.31: Graph of \(y=|f''(x)|=\frac14x^{-3/2}\) for \(f(x)=\sqrt x\).

   In Example 2, the error in \(T_6\) is approximately twice as large as the error in \(M_6\). In practice, this is often the case.

2. The exact value is \({\int_1^4 \sqrt{x}\, dx= \frac23 x^{3/2}\big|_1^4 = \frac{14}{3}}\), so the actual errors are \[ \begin{alignat*}{3} \text{Error}(T_6) & \approx&\: \left\vert \frac{14}3 - 4.661488 \right\vert&\approx 0.00518&\quad\textrm{(less than error bound \(0.0156\))}\\ \text{Error}(M_6) & \approx&\: \left\vert \frac{14}3 - 4.669245\right\vert&\approx 0.00258&\quad\textrm{(less than error bound \(0.0078\))} \end{alignat*} \]

   The actual errors are less than the error bound, so Theorem 1 is verified.

EXAMPLE 3 Obtaining the Desired Accuracy

Find \(N\) such that \(T_N\) approximates \(\int_0^3 e^{-x^2}\, dx\) with an error of at most \(10^{-4}\).

Solution Let \(f(x)=e^{-x^2}\). To apply the error bound, we must find a number \(K_2\) such that \(|f''(x)|\le K_2\) for all \(x\in [0,3]\). We have \(f'(x)=-2xe^{-x^2}\) and \[ f''(x) = (4x^2-2)e^{-x^2} \]

Let's use a graphing utility to plot \(f''(x)\) (Figure 7.32). The graph shows that the maximum value of \(|f''(x)|\) on \([0,3]\) is \(|f''(0)|=|-2|=2\), so we take \(K_2=2\) in the error bound: \[ \text{Error}(T_N) \le \frac{K_2(b-a)^3}{12N^2} = \frac{2(3-0)^3}{12 N^2} = \frac{9}{2N^2} \]

The error is at most \(10^{-4}\) if \[ \frac{9}{2N^2} \le 10^{-4}\quad\Rightarrow\quad N^2\ge \frac{9\times 10^4}{2}\quad\Rightarrow\quad N\ge \frac{300}{\sqrt{2}}\approx 212.1 \]

Figure 7.32: Graph of the second derivative of \(f(x)=e^{-x^2}\).

We conclude that \(T_{213}\) has error at most \(10^{-4}\). We can confirm this using a computer algebra system. A CAS shows that \(T_{213} \approx 0.886207\), whereas the value of the integral to nine places is \(0.886207348\). Thus the error is less than \(10^{-6}\).

Pattern of coefficients in \(S_N\): \[ 1, 4, 2, 4, 2, 4, \ldots, 4, 2, 4, 1 \]

The intermediate coefficients alternate \(4, 2, 4, 2,\dots , 2, 4\) (beginning and ending with \(4\)).

Simpson's Rule

For \(N\) even, the \(N\)th approximation to \(\displaystyle\int_a^b f(x)\, dx\) by Simpson's Rule is \[ S_N = \frac13\Delta x \bigl[y_0 + 4y_1 + 2y_2 + \cdots + 4y_{N-3} + 2y_{N-2} + 4y_{N-1} + y_N \bigr]\tag{3} \]

where \(\Delta x = \dfrac{b-a}{N}\) and \(y_j=f(a+j\,\Delta x)\).

CONCEPTUAL INSIGHT

Both \(T_N\) and \(M_N\) give the exact value of the integral for all \(N\) when \(f(x)\) is a linear function (Exercise 59). However, of all combinations of \(T_{N/2}\) and \(M_{N/2}\), only the particular combination \(S_N = \frac13T_{N/2}+\frac23M_{N/2}\) gives the exact value for all quadratic polynomials (Exercises 60 and 61). In fact, \(S_N\) is also exact for all cubic polynomials (Exercise 62).

EXAMPLE 4

Use Simpson's Rule with \(N = 8\) to approximate \(\displaystyle\int_2^4 \sqrt{1+x^3} \, dx\).

Solution We have \(\Delta x = \tfrac{4-2}8=\tfrac14\). Figure 7.35 shows the endpoints and coefficients needed to compute \(S_8\) using Eq. (3): \[ \begin{align*} \frac13\left(\frac14\right)& \bigl[\sqrt{1+2^3} + 4\sqrt{1+2.25^3} + 2\sqrt{1+2.5^3} + 4\sqrt{1+2.75^3} + 2\sqrt{1+3^3} \\ &+ 4\sqrt{1+3.25^3} + 2\sqrt{1+3.5^3} + 4\sqrt{1+3.75^3} +\sqrt{1+4^3} \bigr] \\ \approx \frac1{12} &\bigl[3 + 4(3.52003) + 2(4.07738) + 4( 4.66871) + 2(5.2915) \\ &+ 4(5.94375)+ 2(6.62382) + 4(7.33037) + 8.06226\bigr] \approx 10.74159 \end{align*} \]

Figure 7.35: Coefficients for \(S_8\) on \([2,4]\) shown above the corresponding endpoint.

EXAMPLE 5 Estimating Integrals from Numerical Data

The velocity (in km/h) of a Piper Cub aircraft traveling due west is recorded every minute during the first 10 minutes after takeoff. Use Simpson's Rule to estimate the distance traveled.

Solution  The distance traveled is the integral of velocity. We convert from minutes to hours because velocity is given in km/h, and thus we apply Simpson's Rule, where the number of intervals is \(N=10\) and each interval has length \(\Delta t = \frac{1}{60}\) hours: \[ \begin{eqnarray*} S_{10} &=& \left(\frac13\right)\left(\frac{1}{60}\right)\big(0\begin{array}[t]{@{}l@{}} {}+4(80)+2(100)+4(128)+2(144)+4(160)\\ {}+2(152)+4(136)+2(128)+4(120)+136\big)\approx 21.2\,\,\textrm{km}\end{array} \end{eqnarray*} \]

The distance traveled is approximately 21.2 km (Figure 7.36).

Figure 7.36: Velocity of a Piper Cub.

THEOREM 2 Error Bound for \(S_N\)

Let \(K_4\) be a number such that \(|f^{(4)}(x)| \le K_4\) for all \(x\in[a,b]\). Then \[ \boxed{\bbox[#FAF8ED,5pt]{\textrm{Error}(S_N) \le \frac{K_4(b- a)^5}{180 N^4}}} \]

Although Simpson's Rule provides good approximations, more sophisticated techniques are implemented in computer algebra systems. These techniques are studied in the area of mathematics called numerical analysis.

EXAMPLE 6

Calculate \(S_8\) for \(\displaystyle\int_1^{3}\dfrac{1}{x}\, dx\). Then:

1. Find a bound for the error in \(S_8\).
2. Find \(N\) such that \(S_N\) has an error of at most \(10^{-6}\).

Solution The width is \(\Delta x = \tfrac{3-1}{8}=\tfrac14\) and the endpoints in the partition of \([1,3]\) are \(1, 1.25, 1.5, \ldots , 2.75, 3\). Using Eq. (3) with \(f(x)=x^{-1}\), we obtain \[ \begin{eqnarray*} S_8 &=& \frac13\left(\frac14\right)\left[ \dfrac{1}{1} + \dfrac{4}{1.25} + \dfrac{2}{1.5} + \dfrac{4}{1.75} + \dfrac{2}{2} + \dfrac{4}{2.25} + \dfrac{2}{2.5} + \dfrac{4}{2.75} + \dfrac{1}{3} \right]\\ &\approx& 1.09873 \end{eqnarray*} \]

1. The fourth derivative \(f^{(4)}(x) = 24x^{-5}\) is decreasing, so the max of \(|f^{(4)}(x)|\) on \([1,3]\) is \(|f^{(4)}(1)| = 24\). Therefore, we use the error bound with \(K_4=24\): \[ \begin{align*} \textrm{Error}(S_N) &\le \frac{K_4(b- a)^5}{180 N^4} =\frac{24(3-1)^5}{180N^4} = \frac{64}{15N^4}\\ \textrm{Error}(S_8) &\le \frac{K_4(b- a)^5}{180 (8)^4} = \frac{24(3-1)^5}{180(8^4)} \approx 0.001 \end{align*} \]

   460

2. The error will be at most \(10^{-6}\) if \(N\) satisfies \[ \mathrm{Error}(S_N) = \frac{64}{15N^4} \le 10^{-6} \]

   In other words, \[ N^4\ge 10^6\left(\dfrac{64}{15}\right)\qquad\hbox{or}\qquad N\ge \left(\dfrac{10^6\cdot 64}{15}\right)^{1/4} \approx 45.45 \]

   Thus, we may take \(N = 46\) (see marginal comment).

Using a CAS, we find that \[ \begin{eqnarray*} S_{46}&\approx& 1.09861241\\ \int_1^{3}\dfrac{1}{x}\, dx= \ln 3 &\approx& 1.09861229 \end{eqnarray*} \]

The error is indeed less than \(10^{-6}\).

GRAPHICAL INSIGHT

Simpson's Rule has an interpretation in terms of parabolas (Figure 7.37). There is a unique parabola passing through the graph of \(f(x)\) at the three points \(x_{2j-2}, x_{2j-1}, x_{2j}\) [Figure 7.37]. On the interval \([x_{2j-2}, x_{2j}]\), the area under the parabola approximates the area under the graph. Simpson's Rule \(S_N\) is equal to the sum of these parabolic approximations (see Exercises 60–61).