Trapezoidal Rule
We know from a previous lesson that we can use Riemann Sums to evaluate a definite integral \(\int\limits_a^b {f\left( x \right)dx}.\)
Riemann Sums use rectangles to approximate the area under a curve.
Another useful integration rule is the Trapezoidal Rule . Under this rule, the area under a curve is evaluated by dividing the total area into little trapezoids rather than rectangles.
Let f (x )[a , b ] . We partition the interval [a , b ] into n equal subintervals, each of width
\[\Delta x = \frac{{b - a}}{n},\]
such that
\[a = {x_0} \lt {x_1} \lt {x_2} \lt \cdots \lt {x_n} = b.\]
Figure 1. The Trapezoidal Rule for approximating \(\int\limits_a^b {f\left( x \right)dx}\) is given by
\[\int\limits_a^b {f\left( x \right)dx} \approx {T_n} = {\frac{{\Delta x}}{2}}\left[ {{f\left( {{x_0}} \right)} + {2f\left( {{x_1}} \right)} + {2f\left( {{x_2}} \right)} + \cdots + {2f\left( {{x_{n - 1}}} \right)} + {f\left( {{x_n}} \right)}} \right],\]
where \(\Delta x = \frac{{b - a}}{n}\) and \({x_i} = a + i\Delta x.\)
As \(n \to \infty,\) the right-hand side of the expression approaches the definite integral \(\int\limits_a^b {f\left( x \right)dx}.\)
Solved Problems
Example 1.
Use the Trapezoidal Rule with \(n = 6\) to approximate \[\int\limits_0^\pi {{{\sin }^2}xdx}.\]
Solution.
Here
\[f\left( x \right) = {\sin ^2}x,\;\; a = 0,\;\; b = \pi .\]
The width of each subinterval is
\[\Delta x = \frac{{b - a}}{n} = \frac{\pi }{6},\]
so the grid points have the coordinates \({x_i} = \frac{{i\pi }}{6}.\)
Calculate the values of the function \(f\left( x \right)\) at the points \({x_i}:\)
\[f\left( {{x_0}} \right) = f\left( 0 \right) = {\sin ^2}0 = {0^2} = 0;\]
\[f\left( {{x_1}} \right) = f\left( {\frac{\pi }{6}} \right) = {\sin ^2}\frac{\pi }{6} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4};\]
\[f\left( {{x_2}} \right) = f\left( {\frac{{2\pi }}{6}} \right) = {\sin ^2}\frac{\pi }{3} = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = \frac{3}{4};\]
\[f\left( {{x_3}} \right) = f\left( {\frac{{3\pi }}{6}} \right) = {\sin ^2}\frac{\pi }{2} = {1^2} = 1;\]
\[f\left( {{x_4}} \right) = f\left( {\frac{{4\pi }}{6}} \right) = {\sin ^2}\frac{{2\pi }}{3} = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = \frac{3}{4};\]
\[f\left( {{x_5}} \right) = f\left( {\frac{{5\pi }}{6}} \right) = {\sin ^2}\frac{{5\pi }}{6} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4};\]
\[f\left( {{x_6}} \right) = f\left( \pi \right) = {\sin ^2}\pi = {0^2} = 0.\]
The Trapezoidal Rule formula is written in the form
\[\int\limits_0^\pi {{{\sin }^2}xdx} \approx {T_6} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + \cdots + 2f\left( {{x_5}} \right) + f\left( {{x_6}} \right)} \right] = \frac{\pi }{{12}}\left[ {0 + 2 \cdot \frac{1}{4} + 2 \cdot \frac{3}{4} + 2 \cdot 1 + 2 \cdot \frac{3}{4} + 2 \cdot \frac{1}{4} + 0} \right] = \frac{\pi }{{12}}\left[ {\frac{1}{2} + \frac{3}{2} + 2 + \frac{3}{2} + \frac{1}{2}} \right] = \frac{\pi }{{12}} \cdot \frac{{12}}{2} = \frac{\pi }{2}.\]
We can also determine the exact value of the integral:
\[\int\limits_0^\pi {{{\sin }^2}xdx} = \frac{1}{2}\int\limits_0^\pi {\left( {1 - \cos 2x} \right)dx} = \frac{1}{2}\left[ {x - \frac{{\sin 2x}}{2}} \right]_0^\pi = \frac{1}{2}\left[ {\left( {\pi - 0} \right) - 0} \right] = \frac{\pi }{2}.\]
So, in this particular example, the trapezoidal approximation \({T_6}\) coincides with the exact value of the integral.
Example 2.
A function \(f\left( x \right)\) is given by the table of values. Approximate the area under the curve \(y = f\left( x \right)\) between \(x = 0\) and \(x = 8\) using the Trapezoidal Rule with \(n = 4\) subintervals.
Solution.
The Trapezoidal Rule formula for \(n= 4\) subintervals has the form
\[{T_4} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 2f\left( {{x_3}} \right) + f\left( {{x_4}} \right)} \right].\]
The width of the subinterval is \(\Delta x = 2.\)
Substituting the values of the function from the table, we find the approximate area under the curve:
\[A \approx {T_4} = \frac{2}{2}\left[ {3 + 2 \cdot 7 + 2 \cdot 11 + 2 \cdot 9 + 3} \right] = 3 + 14 + 22 + 18 + 3 = 60.\]
Example 3.
A function \(f\left( x \right)\) is given by the table of values. Approximate the area under the curve \(y = f\left( x \right)\) between \(x = -4\) and \(x = 2\) using the Trapezoidal Rule with \(n = 6\) subintervals.
Solution.
We apply the Trapezoidal Rule formula with \(n = 6\) subintervals which is given by
\[{T_6} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 2f\left( {{x_3}} \right) + 2f\left( {{x_4}} \right) + 2f\left( {{x_5}} \right) + f\left( {{x_6}} \right)} \right].\]
The width of each interval is \(\Delta x = 1.\)
The function values are known from the table, so we can easily calculate the approximate value of the area:
\[A \approx {T_6} = \frac{1}{2}\left[ {0 + 2 \cdot 4 + 2 \cdot 5 + 2 \cdot 3 + 2 \cdot 10 + 2 \cdot 11 + 2} \right] = \frac{1}{2}\left[ {8 + 10 + 6 + 20 + 22 + 2} \right] = \frac{{68}}{2} = 34.\]
Example 4.
Approximate the area under the curve \(y = f\left( x \right)\) between \(x = 0\) and \(x = 10\) using the Trapezoidal Rule with \(n = 5\) subintervals.
Figure 2. Solution.
The Trapezoidal Rule formula for \(n = 5\) intervals is given by
\[{T_5} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 2f\left( {{x_3}} \right) + 2f\left( {{x_4}} \right) + f\left( {{x_5}} \right)} \right].\]
It follows from the figure that \(\Delta x = 2.\) The function values at the endpoints of the intervals are
\[f\left( {{x_0}} \right) = f\left( 0 \right) = 4;\]
\[f\left( {{x_1}} \right) = f\left( 2 \right) = 6;\]
\[f\left( {{x_2}} \right) = f\left( 4 \right) = 6;\]
\[f\left( {{x_3}} \right) = f\left( 6 \right) = 4;\]
\[f\left( {{x_4}} \right) = f\left( 8 \right) = 4;\]
\[f\left( {{x_5}} \right) = f\left( {10} \right) = 5.\]
Hence, the approximate value of the area is equal to
\[A \approx {T_5} = \frac{2}{2}\left[ {4 + 2 \cdot 6 + 2 \cdot 6 + 2 \cdot 4 + 2 \cdot 4 + 5} \right] = 49.\]
Example 5.
Approximate the area under the curve \[y = {2^x}\] between \(x = -1\) and \(x = 3\) using the Trapezoidal Rule with \(n = 4\) subintervals.
Solution.
Figure 3. The Trapezoidal Rule formula for \(n = 4\) has the form
\[{T_4} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 2f\left( {{x_3}} \right) + f\left( {{x_4}} \right)} \right].\]
Compute the function values \(f\left( {{x_i}} \right):\)
\[f\left( {{x_0}} \right) = f\left( { - 1} \right) = {2^{ - 1}} = \frac{1}{2};\]
\[f\left( {{x_1}} \right) = f\left( 0 \right) = {2^0} = 1;\]
\[f\left( {{x_2}} \right) = f\left( 1 \right) = {2^1} = 2;\]
\[f\left( {{x_3}} \right) = f\left( 2 \right) = {2^2} = 4;\]
\[f\left( {{x_4}} \right) = f\left( 3 \right) = {2^3} = 8.\]
As \(\Delta x = 1,\) we get
\[A \approx {T_4} = \frac{1}{2}\left[ {\frac{1}{2} + 2 \cdot 1 + 2 \cdot 2 + 2 \cdot 4 + 8} \right] = \frac{1}{2} \cdot 22\frac{1}{2} = 11\frac{1}{4}.\]
Example 6.
Approximate the area under the curve \[y = \frac{1}{x}\] between \(x = 1\) and \(x = 5\) using the Trapezoidal Rule with \(n = 4\) subintervals.
Solution.
Figure 4. We write the Trapezoidal Rule formula for \(n = 4\) subintervals:
\[{T_4} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 2f\left( {{x_3}} \right) + f\left( {{x_4}} \right)} \right].\]
The function has the following values at the points \({x_i}:\)
\[f\left( {{x_0}} \right) = f\left( 1 \right) = \frac{1}{1} = 1;\]
\[f\left( {{x_1}} \right) = f\left( 2 \right) = \frac{1}{2};\]
\[f\left( {{x_2}} \right) = f\left( 3 \right) = \frac{1}{3};\]
\[f\left( {{x_3}} \right) = f\left( 4 \right) = \frac{1}{4};\]
\[f\left( {{x_4}} \right) = f\left( 5 \right) = \frac{1}{5}.\]
Since \(\Delta x = 1,\) we obtain
\[A \approx {T_4} = \frac{1}{2}\left[ {1 + 2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{3} + 2 \cdot \frac{1}{4} + \frac{1}{5}} \right] = \frac{1}{2}\left[ {1 + 1 + \frac{2}{3} + \frac{1}{2} + \frac{1}{5}} \right] = \frac{1}{2} \cdot \frac{{30 + 30 + 20 + 15 + 8}}{{30}} = \frac{1}{2} \cdot \frac{{101}}{{30}} = \frac{{101}}{{60}}\]
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