Calculus

Integration of Functions

Integration of Functions Logo

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 \(f\left( x \right)\) be continuous on \(\left[ {a,b} \right].\) We partition the interval \(\left[ {a,b} \right]\) into \(n\) equal subintervals, each of width

    \[{\Delta x = \frac{{b – a}}{n},}\]

    such that

    \[a = {x_0} \lt {x_1} \lt {x_2} \lt \cdots \lt {x_n} = b.\]

    Trapezoidal Rule Concept
    Figure 1.

    The Trapezoidal Rule for approximating \(\int\limits_a^b {f\left( x \right)dx}\) is given by

    \[{\int\limits_a^b {f\left( x \right)dx} \approx {T_n} }={ {\frac{{\Delta x}}{2}}\left[ {{f\left( {{x_0}} \right)} + {2f\left( {{x_1}} \right)} }\right.}+{\left.{ {2f\left( {{x_2}} \right)} + \cdots }\right.}+{\left.{ {2f\left( {{x_{n – 1}}} \right)} + {f\left( {{x_n}} \right)}} \right],}\]

    where \(\Delta x = \large{\frac{{b – a}}{n}}\normalsize\) 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

    Click a problem to see the solution.

    Example 1

    Use the Trapezoidal Rule with \(n = 6\) to approximate \(\int\limits_0^\pi {{{\sin }^2}xdx}.\)

    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.

    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.

    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.

    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.

    Example 6

    Approximate the area under the curve \(y = \large{\frac{1}{x}}\normalsize\) between \(x = 1\) and \(x = 5\) using the Trapezoidal Rule with \(n = 4\) subintervals.

    Example 7

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

    Example 8

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

    Example 9

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

    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.

    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.

    Example 1.

    Use the Trapezoidal Rule with \(n = 6\) to approximate \(\int\limits_0^\pi {{{\sin }^2}xdx}.\)

    Solution.

    Here

    \[{f\left( x \right) = {\sin ^2}x,\;\;}\kern0pt{a = 0,\;\;}\kern0pt{b = \pi .}\]

    The width of each subinterval is

    \[{\Delta x = \frac{{b – a}}{n} = \frac{\pi }{6},}\]

    so the grid points have the coordinates \({x_i} = \large{\frac{{i\pi }}{6}}\normalsize.\)

    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

    \[{\int\limits_0^\pi {{{\sin }^2}xdx} \approx {T_6} }={ \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + \cdots }\right.}+{\left.{ 2f\left( {{x_5}} \right) + f\left( {{x_6}} \right)} \right] }={ \frac{\pi }{{12}}\left[ {0 + 2 \cdot \frac{1}{4} + 2 \cdot \frac{3}{4} }\right.}+{\left.{ 2 \cdot 1 + 2 \cdot \frac{3}{4} + 2 \cdot \frac{1}{4} + 0} \right] }={ \frac{\pi }{{12}}\left[ {\frac{1}{2} + \frac{3}{2} + 2 + \frac{3}{2} + \frac{1}{2}} \right] }={ \frac{\pi }{{12}} \cdot \frac{{12}}{2} }={ \frac{\pi }{2}}\]

    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.
    A function given by the table

    Solution.

    The Trapezoidal Rule formula for \(n= 4\) subintervals has the form

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

    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:

    \[{A \approx {T_4} \text{ = }}\kern0pt{ \frac{2}{2}\left[ {3 + 2 \cdot 7 + 2 \cdot 11 + 2 \cdot 9 + 3} \right] }={ 3 + 14 + 22 + 18 + 3 }={ 60.}\]

    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.
    Function given by the table

    Solution.

    We apply the Trapezoidal Rule formula with \(n = 6\) subintervals which is given by

    \[{{T_6} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 2f\left( {{x_3}} \right) }\right.}+{\left.{ 2f\left( {{x_4}} \right) + 2f\left( {{x_5}} \right) }\right.}+{\left.{ f\left( {{x_6}} \right)} \right].}\]

    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:

    \[{A \approx {T_6} \text{ = }}\kern0pt{\frac{1}{2}\left[ {0 + 2 \cdot 4 + 2 \cdot 5 }\right.}+{\left.{ 2 \cdot 3 + 2 \cdot 10 }\right.}+{\left.{ 2 \cdot 11 + 2} \right] }={ \frac{1}{2}\left[ {8 + 10 + 6 }\right.}+{\left.{ 20 + 22 + 2} \right] }={ \frac{{68}}{2} }={ 34.}\]

    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.
    Trapezoidal approximation with n=5 intervals
    Figure 2.

    Solution.

    The Trapezoidal Rule formula for \(n = 5\) intervals is given by

    \[{{T_5} = \frac{{\Delta x}}{2}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 2f\left( {{x_3}} \right) }\right.}+{\left.{ 2f\left( {{x_4}} \right) + f\left( {{x_5}} \right)} \right].}\]

    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

    \[{A \approx {T_5} }={ \frac{2}{2}\left[ {4 + 2 \cdot 6 }\right.}+{\left.{ 2 \cdot 6 + 2 \cdot 4 }\right.}+{\left.{ 2 \cdot 4 + 5} \right] }={ 49.}\]

    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.

    Trapezoidal approximation of the function y=2^x with n=5 intervals
    Figure 3.

    The Trapezoidal Rule formula for \(n = 4\) has the form

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

    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

    \[{A \approx {T_4} \text{ = }}\kern0pt{ \frac{1}{2}\left[ {\frac{1}{2} + 2 \cdot 1 + 2 \cdot 2 + 2 \cdot 4 + 8} \right] }={ \frac{1}{2} \cdot 22\frac{1}{2} }={ 11\frac{1}{4}.}\]

    Example 6.

    Approximate the area under the curve \(y = \large{\frac{1}{x}}\normalsize\) between \(x = 1\) and \(x = 5\) using the Trapezoidal Rule with \(n = 4\) subintervals.

    Solution.

    Area under the curve y=1/x
    Figure 4.

    We write the Trapezoidal Rule formula for \(n = 4\) subintervals:

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

    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

    \[{A \approx {T_4} }={ \frac{1}{2}\left[ {1 + 2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{3} }\right.}+{\left.{ 2 \cdot \frac{1}{4} + \frac{1}{5}} \right] }={ \frac{1}{2}\left[ {1 + 1 + \frac{2}{3} }\right.}+{\left.{ \frac{1}{2} + \frac{1}{5}} \right] }={ \frac{1}{2} \cdot \frac{{30 + 30 + 20 + 15 + 8}}{{30}} }={ \frac{1}{2} \cdot \frac{{101}}{{30}} }={ \frac{{101}}{{60}}}\]

    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

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

    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:

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

    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 \(\int\limits_0^1 {\large{\frac{{dx}}{{1 + {x^2}}}}\normalsize} \) and calculate the approximate value of \(\pi.\) Round the answer to \(2\) decimal places.

    Solution.

    We evaluate the given integral by the formula

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

    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:

    The values of the function f(x)=1/(1+x^2)
    Figure 5.

    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

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

    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

    \[{\left| \varepsilon \right| }={ \frac{{4.5 – 4}}{{4.5}} }={ \frac{{0.5}}{{4.5}} }={ \frac{1}{9} }\approx{ 11.1\% }\]

    The final answer is

    \[{A \approx 4,\;}\kern0pt{\left| \varepsilon \right| = 11.1\%}\]

    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

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

    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:

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

    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

    \[{\left| \varepsilon \right| = \frac{{2 – 1.896}}{2} }={ 0.052 }={ 5.2\%} \]

    The answer is

    \[{A \approx 1.896,\;}\kern0pt{\left| \varepsilon \right| = 5.2\%}\]