# Trapezoidal Rule

## Solved Problems

### Example 7.

Approximate the integral $\int\limits_0^1 {{x^3}dx}$ using the Trapezoidal Rule with $$n = 2$$ subintervals.

Solution.

The Trapezoidal Rule formula with $$n = 2$$ subintervals is written as

${T_2} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + f\left( {{x_2}} \right)} \right],$

where

$\Delta x = \frac{{b - a}}{n} = \frac{{1 - 0}}{2} = \frac{1}{2},$

and the function has the following values:

$f\left( {{x_0}} \right) = f\left( 0 \right) = {0^3} = 0;$
$f\left( {{x_1}} \right) = f\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{8};$
$f\left( {{x_2}} \right) = f\left( 1 \right) = {1^3} = 1.$

Hence

$\int\limits_0^1 {{x^3}dx} \approx {T_2} = \frac{1}{4}\left[ {0 + 2 \cdot \frac{1}{8} + 1} \right] = \frac{1}{4} \cdot \frac{5}{4} = \frac{5}{{16}}$

### Example 8.

Approximate the integral $\int\limits_0^2 {{x^2}dx}$ using the Trapezoidal Rule with $$n = 3$$ subintervals.

Solution.

We write the Trapezoidal Rule formula with $$n = 3$$ subintervals:

${T_3} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + f\left( {{x_3}} \right)} \right],$

where

$\Delta x = \frac{{b - a}}{n} = \frac{{2 - 0}}{3} = \frac{2}{3},$

and the function has the following values at $${x_i}:$$

$f\left( {{x_0}} \right) = f\left( 0 \right) = {0^2} = 0;$
$f\left( {{x_1}} \right) = f\left( {\frac{2}{3}} \right) = {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9};$
$f\left( {{x_2}} \right) = f\left( {\frac{4}{3}} \right) = {\left( {\frac{4}{3}} \right)^2} = \frac{{16}}{9};$
$f\left( {{x_3}} \right) = f\left( 2 \right) = {2^2} = 4.$

Then

$\int\limits_0^2 {{x^2}dx} \approx {T_3} = \frac{1}{3}\left[ {0 + 2 \cdot \frac{4}{9} + 2 \cdot \frac{{16}}{9} + 4} \right] = \frac{1}{3}\left[ {\frac{8}{9} + \frac{{32}}{9} + 4} \right] = \frac{1}{3} \cdot \frac{{8 + 32 + 36}}{9} = \frac{1}{3} \cdot \frac{{76}}{9} = \frac{{76}}{{27}}$

### Example 9.

Using the Trapezoidal Rule with $$n = 10$$ subintervals, evaluate the integral $\int\limits_0^1 {\frac{{dx}}{{1 + {x^2}}}}$ and calculate the approximate value of $$\pi.$$ Round the answer to $$2$$ decimal places.

Solution.

We evaluate the given integral by the formula

${T_{10}} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + \cdots + 2f\left( {{x_9}} \right) + f\left( {{x_{10}}} \right)} \right].$

The width of each subinterval is

$\Delta x = \frac{{b - a}}{n} = \frac{{1 - 0}}{{10}} = 0.1$

The function values at the points $${x_i}$$ are given below:

Plugging in the values from the table into our equation, we obtain:

$\int\limits_0^1 {\frac{{dx}}{{1 + {x^2}}}} \approx {T_{10}} = 0.05 \times 15.6996 = 0.7850$

Let's also evaluate this integral by the Fundamental Theorem of Calculus:

$\int\limits_0^1 {\frac{{dx}}{{1 + {x^2}}}} = \left[ {\arctan x} \right]_0^1 = \arctan 1 - \arctan 0 = \frac{\pi }{4} - 0 = \frac{\pi }{4}.$

Hence, the approximate value of $$\pi$$ with an accuracy of two decimal places is

$\pi \approx 4\int\limits_0^1 {\frac{{dx}}{{1 + {x^2}}}} = 4 \times 0.7850 = 3.14$

### Example 10.

Using the Trapezoidal Rule with $$n = 3$$ subintervals approximate the area under the curve $f\left( x \right) = 3x - {x^2}$ between $$x = 0$$ and $$x = 3.$$ Estimate the relative percent error of the approximation.

Solution.

The Trapezoidal Rule with $$n = 3$$ segments is written in the form

${T_3} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + f\left( {{x_3}} \right)} \right].$

The width of the subinterval is equal to

$\Delta x = \frac{{b - a}}{n} = \frac{{3 - 0}}{3} = 1.$

We calculate the values of the function at the points $${x_i}:$$

$f\left( {{x_0}} \right) = f\left( 0 \right) = 3 \cdot 0 - {0^2} = 0;$
$f\left( {{x_1}} \right) = f\left( 1 \right) = 3 \cdot 1 - {1^2} = 2;$
$f\left( {{x_2}} \right) = f\left( 2 \right) = 3 \cdot 2 - {2^2} = 2;$
$f\left( {{x_3}} \right) = f\left( 3 \right) = 3 \cdot 3 - {3^2} = 0.$

Hence, the approximate value of the area under the curve is given by

$A \approx {T_3} = \frac{1}{2}\left[ {0 + 2 \cdot 2 + 2 \cdot 2 + 0} \right] = 4.$

The true solution is found by integration:

$\int\limits_0^3 {f\left( x \right)dx} = \int\limits_0^3 {\left( {3x - {x^2}} \right)dx} = \left[ {\frac{{3{x^2}}}{2} - \frac{{{x^3}}}{3}} \right]_0^3 = \frac{{27}}{2} - \frac{{27}}{3} - 0 = \frac{9}{2} = 4.5$

So the relative error is

$\left| \varepsilon \right| = \frac{{4.5 - 4}}{{4.5}} = \frac{{0.5}}{{4.5}} = \frac{1}{9} \approx 11.1\%$

$A \approx 4,\;\left| \varepsilon \right| = 11.1\%$

### Example 11.

Using the Trapezoidal Rule with $$n = 4$$ subintervals approximate the area under the sine curve $f\left( x \right) = \sin x$ between $$x = 0$$ and $$x = \pi$$ to $$3$$ decimal places. Estimate the relative percent error of the approximation.

Solution.

The Trapezoidal Rule formula with $$n = 4$$ segments is given by

${T_4} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 2f\left( {{x_3}} \right) + f\left( {{x_4}} \right)} \right].$

Determine the width of each subinterval:

$\Delta x = \frac{{b - a}}{n} = \frac{{\pi - 0}}{4} = \frac{\pi }{4}.$

Calculate the values of the sine function:

$f\left( {{x_0}} \right) = f\left( 0 \right) = \sin 0 = 0;$
$f\left( {{x_1}} \right) = f\left( {\frac{\pi }{4}} \right) = \sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{2};$
$f\left( {{x_2}} \right) = f\left( {\frac{\pi }{2}} \right) = \sin \frac{\pi }{2} = 1;$
$f\left( {{x_3}} \right) = f\left( {\frac{{3\pi }}{4}} \right) = \sin \frac{{3\pi }}{4} = \frac{{\sqrt 2 }}{2};$
$f\left( {{x_4}} \right) = f\left( \pi \right) = \sin \pi = 0.$

Substituting these values, we find the approximate value of the area:

$A \approx {T_4} = \frac{\pi }{8}\left[ {0 + 2 \cdot \frac{{\sqrt 2 }}{2} + 2 \cdot 1 + 2 \cdot \frac{{\sqrt 2 }}{2} + 0} \right] = \frac{\pi }{8}\left( {2\sqrt 2 + 2} \right) = \frac{\pi }{4}\left( {\sqrt 2 + 1} \right) \approx 1.896$

Integrating the sine function we get the true solution:

$\int\limits_0^\pi {f\left( x \right)dx} = \int\limits_0^\pi {\sin xdx} = \left[ { - \cos x} \right]_0^\pi = - \cos \pi + \cos 0 = - \left( { - 1} \right) + 1 = 2.$

Thus, the relative error is

$\left| \varepsilon \right| = \frac{{2 - 1.896}}{2} = 0.052 = 5.2\%$

$A \approx 1.896,\;\left| \varepsilon \right| = 5.2\%$