Calculus

Integration of Functions

Integration of Functions Logo

Simpson’s Rule

  • Simpson’s Rule is a numerical method that approximates the value of a definite integral by using quadratic functions.

    This method is named after the English mathematician Thomas Simpson \(\left( {1710 – 1761} \right).\)

    Simpson’s Rule is based on the fact that given three points, we can find the equation of a quadratic through those points.

    To obtain an approximation of the definite integral \(\int\limits_a^b {f\left( x \right)dx} \) using Simpson’s Rule, we partition the interval \(\left[ {a,b} \right]\) into an even number \(n\) of subintervals, each of width

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

    On each pair of consecutive subintervals \(\left[ {{x_{i – 1}},{x_i}} \right],\) \(\left[ {{x_i},{x_{i + 1}}} \right],\) we consider a quadratic function \(y = a{x^2} + bx + c\) such that it passes through the points \(\left( {{x_{i – 1}},f\left( {{x_{i – 1}}} \right)} \right),\) \(\left( {{x_i},f\left( {{x_i}} \right)} \right),\) \(\left( {{x_{i + 1}},f\left( {{x_{i + 1}}} \right)} \right).\)

    Simpson's method of integration
    Figure 1.

    If the function \(f\left( x \right)\) is continuous on \(\left[ {a,b} \right],\) then

    \[{\int\limits_a^b {f\left( x \right)dx} }\approx{ {\frac{{\Delta x}}{3}}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) }\right.}+{\left.{ 2f\left( {{x_4}} \right) + \cdots }\right.}+{\left.{ 4f\left( {{x_{n – 1}}} \right) + f\left( {{x_n}} \right)} \right].}\]

    The coefficients in Simpson’s Rule have the following pattern:

    \[{\underbrace {1,4,2,4,2, \ldots ,4,2,4,1}_{{n + 1}\;\text{points}}.}\]


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Use Simpson’s Rule with \(n = 4\) to approximate the integral \(\int\limits_0^8 {\sqrt x dx}.\)

    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 \(x = 4\) using Simpson’s Rule with \(n = 4\) subintervals.

    Example 3

    A function \(f\left( x \right)\) is given as a table of values. Approximate the area under the curve \(y = f\left( x \right)\) from \(x = -4\) and \(x = 8\) using Simpson’s Rule with \(n = 6\) subintervals.

    Example 4

    Approximate the area under the curve \(y = f\left( x \right)\) between \(x = -1\) and \(x = 5\) using Simpson’s Rule with \(n = 6\) subintervals.

    Example 5

    Approximate the area under the curve \(y = {3^x}\) between \(x = -2\) and \(x = 2\) using Simpson’s Rule with \(n = 4\) subintervals.

    Example 6

    Approximate the integral \(\int\limits_1^2 {\large{\frac{{dx}}{x}}\normalsize}\) using Simpson’s Rule with \(n = 2\) subintervals.

    Example 7

    Approximate the integral \(\int\limits_0^2 {{x^3}dx}\) using Simpson’s Rule with \(n = 4\) subintervals.

    Example 8

    Using Simpson’s Rule with \(n = 4\) subintervals, evaluate the integral \(\int\limits_0^1 {{e^x}dx}.\) Round the answer to \(3\) decimal places.

    Example 9

    Using Simpson’s Rule with \(n = 2\) subintervals, evaluate the integral \(\int\limits_1^3 {\ln xdx}.\) Round the answer to \(3\) decimal places.

    Example 10

    Using Simpson’s Rule with \(n = 4\) subintervals approximate the area under the sine curve \(f\left( x \right) = \sin x\) between \(x = 0\) and \(x = \pi.\)

    Example 11

    Using Simpson’s Rule with \(n = 4\) subintervals approximate the area under the inverse cosine curve \(f\left( x \right) = \arccos x\) between \(x = -1\) and \(x = 1.\)

    Example 1.

    Use Simpson’s Rule with \(n = 4\) to approximate the integral \(\int\limits_0^8 {\sqrt x dx}.\)

    Solution.

    It is easy to see that the width of each subinterval is

    \[{\Delta x = \frac{{b – a}}{n} = \frac{{8 – 0}}{4} = 2,}\]

    and the endpoints \({x_i}\) have coordinates

    \[{x_i} = \left\{ {0,2,4,6,8} \right\}.\]

    Calculate the function values at the points \({x_i}:\)

    \[{f\left( {{x_0}} \right) = f\left( 0 \right) = \sqrt 0 = 0;}\]

    \[{f\left( {{x_1}} \right) = f\left( 2 \right) = \sqrt 2 ;}\]

    \[{f\left( {{x_2}} \right) = f\left( 4 \right) = \sqrt 4 = 2;}\]

    \[{f\left( {{x_3}} \right) = f\left( 6 \right) = \sqrt 6 ;}\]

    \[{f\left( {{x_4}} \right) = f\left( 8 \right) = \sqrt 8 = 2\sqrt 2 .}\]

    Substitute all these values into the Simpson’s Rule formula:

    \[{\int\limits_0^8 {\sqrt x dx} }\approx{ \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) }\right.}+{\left.{ f\left( {{x_4}} \right)} \right] }={ \frac{2}{3}\left[ {0 + 4 \cdot \sqrt 2 + 2 \cdot 2 }\right.}+{\left.{ 4 \cdot \sqrt 6 + 2\sqrt 2 } \right] }={ \frac{2}{3}\left[ {6\sqrt 2 + 4 + 4\sqrt 6 } \right] }\approx{ 14.86}\]

    The true solution for the integral is

    \[{\int\limits_0^8 {\sqrt x dx} }={ \int\limits_0^8 {{x^{\frac{1}{2}}}dx} }={ \left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^8 }={ \frac{2}{3}\left[ {\sqrt {{x^3}} } \right]_0^8 }={ \frac{2}{3}\sqrt {{8^3}} }={ \frac{2}{3}\sqrt {{2^9}} }={ \frac{2}{3} \cdot 16\sqrt 2 }={ \frac{{32\sqrt 2 }}{3} }\approx{ 15.08}\]

    Hence, the error in approximating the integral is

    \[{\left| \varepsilon \right| = \left| {\frac{{15.08 – 14.86}}{{15.08}}} \right| }\approx{ 0.015 }={ 1.5\%} \]

    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 \(x = 4\) using Simpson’s Rule with \(n = 4\) subintervals.
    A function given by the table

    Solution.

    For \(n= 4\) subintervals, Simpson’s rule is given by the following equation:

    \[{{S_4} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + f\left( {{x_4}} \right)} \right].}\]

    The width of the subinterval is

    \[\Delta x = \frac{{b – a}}{n} = \frac{{4 – 0}}{4} = 1.\]

    Substitute the values of the function from the table and calculate the approximate value of the area under the curve:

    \[{A = {S_4} }\approx{ \frac{1}{3}\left[ {2 + 4 \cdot 7 + 2 \cdot 12 + 4 \cdot 10 + 5} \right] }={ \frac{1}{3}\left[ {2 + 28 + 24 + 40 + 5} \right] }={ \frac{1}{3} \cdot 99 }={ 33}\]

    Example 3.

    A function \(f\left( x \right)\) is given as a table of values. Approximate the area under the curve \(y = f\left( x \right)\) from \(x = -4\) and \(x = 8\) using Simpson’s Rule with \(n = 6\) subintervals.
    A function given as a table of values

    Solution.

    We use Simpson’s rule formula which has the following form for \(n = 6\) subintervals:

    \[{{S_6} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + 2f\left( {{x_4}} \right) }\right.}+{\left.{ 4f\left( {{x_5}} \right) + f\left( {{x_6}} \right)} \right].}\]

    Determine the width \(\Delta x\) of the subinterval:

    \[\Delta x = \frac{{b – a}}{n} = \frac{{8 – \left( { – 4} \right)}}{6} = 2.\]

    Using the values of the function given in the table, we calculate the approximate value of the area under the curve.

    \[{A = {S_6} \approx \frac{2}{3}\left[ {1 + 4 \cdot 3 + 2 \cdot 4 }\right.}+{\left.{ 4 \cdot 4 + 2 \cdot 6 }\right.}+{\left.{ 4 \cdot 9 + 14} \right] }={ \frac{2}{3}\left[ {1 + 12 + 8 + 16 }\right.}+{\left.{ 12 + 36 + 14} \right] }={ \frac{2}{3} \cdot 99 }={ 66}\]

    Example 4.

    Approximate the area under the curve \(y = f\left( x \right)\) between \(x = -1\) and \(x = 5\) using Simpson’s Rule with \(n = 6\) subintervals.

    Solution.

    Area under the curve y=f(x)
    Figure 2.

    The Simpson’s Rule formula for \(n = 6\) subintervals is given by

    \[{{S_6} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + 2f\left( {{x_4}} \right) }\right.}+{\left.{ 4f\left( {{x_5}} \right) + f\left( {{x_6}} \right)} \right].}\]

    It follows from the figure that \(\Delta x = 1.\) The function values at the endpoints of the intervals are

    \[f\left( {{x_0}} \right) = f\left( -1 \right) = 4;\]

    \[f\left( {{x_1}} \right) = f\left( 0 \right) = 3;\]

    \[f\left( {{x_2}} \right) = f\left( 1 \right) = 2;\]

    \[f\left( {{x_3}} \right) = f\left( 2 \right) = 3;\]

    \[f\left( {{x_4}} \right) = f\left( 3 \right) = 6;\]

    \[f\left( {{x_5}} \right) = f\left( 4 \right) = 6;\]

    \[f\left( {{x_6}} \right) = f\left( 5 \right) = 4.\]

    So, the approximate value of the area under the curve is

    \[{A = {S_6} \approx \frac{1}{3}\left[ {4 + 4 \cdot 3 + 2 \cdot 2 }\right.}+{\left.{ 4 \cdot 3 + 2 \cdot 6 }\right.}+{\left.{ 4 \cdot 6 + 4} \right] }={ \frac{1}{3}\left[ {4 + 12 + 4 + 12 }\right.}+{\left.{ 12 + 24 + 4} \right] }={ \frac{1}{3} \cdot 72 }={ 24}\]

    Example 5.

    Approximate the area under the curve \(y = {3^x}\) between \(x = -2\) and \(x = 2\) using Simpson’s Rule with \(n = 4\) subintervals.

    Solution.

    Area under the curve y=3^x
    Figure 3.

    The Simpson’s Rule formula formula for \(n = 4\) is written in the form

    \[{{S_4} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) }\right.}+{\left.{ f\left( {{x_4}} \right)} \right].}\]

    We calculate the function values \(f\left( {{x_i}} \right):\)

    \[f\left( {{x_0}} \right) = f\left( {-2} \right) = {3^{ – 2}} = \frac{1}{9};\]

    \[f\left( {{x_1}} \right) = f\left( {-1} \right) = {3^{ – 1}} = \frac{1}{3};\]

    \[f\left( {{x_2}} \right) = f\left( 0 \right) = {3^0} = 1;\]

    \[f\left( {{x_3}} \right) = f\left( 1 \right) = {3^1} = 3;\]

    \[f\left( {{x_4}} \right) = f\left( 2 \right) = {3^2} = 9.\]

    As \(\Delta x = 1,\) we obtain

    \[{A = {S_4} }\approx{ \frac{1}{3}\left[ {\frac{1}{9} + 4 \cdot \frac{1}{3} + 2 \cdot 1 }\right.}+{\left.{ 4 \cdot 3 + 9} \right] }={ \frac{1}{3}\left[ {\frac{1}{9} + \frac{4}{3} + 23} \right] }={ \frac{1}{3} \cdot \frac{{1 + 12 + 207}}{9} }={ \frac{{220}}{{27}} }={ 8\frac{4}{{27}}.}\]

    Example 6.

    Approximate the integral \(\int\limits_1^2 {\large{\frac{{dx}}{x}}\normalsize}\) using Simpson’s Rule with \(n = 2\) subintervals.

    Solution.

    The Simpson’s Rule formula with \(n = 2\) subintervals is given by

    \[{{S_2} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ f\left( {{x_2}} \right)} \right].}\]

    The width of the subinterval is

    \[{\Delta x = \frac{{b – a}}{n} }={ \frac{{2 – 1}}{2} }={ \frac{1}{2}.}\]

    Calculate the values of the function at the points \({x_i} = \left\{ {1,\large{\frac{3}{2}}\normalsize,2} \right\}:\)

    \[f\left( {{x_0}} \right) = f\left( 1 \right) = \frac{1}{1} = 1;\]

    \[f\left( {{x_1}} \right) = f\left( {\frac{3}{2}} \right) = \frac{1}{{\frac{3}{2}}} = \frac{2}{3};\]

    \[f\left( {{x_2}} \right) = f\left( 2 \right) = \frac{1}{2}.\]

    Then

    \[{\int\limits_1^2 {\frac{{dx}}{x}} \approx {S_2} \text{ = }}\kern0pt{ \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + f\left( {{x_2}} \right)} \right] }={ \frac{1}{6}\left[ {1 + 4 \cdot \frac{2}{3} + \frac{1}{2}} \right] }={ \frac{1}{6} \cdot \frac{{6 + 16 + 3}}{6} }={ \frac{{25}}{{36}}.}\]

    Example 7.

    Approximate the integral \(\int\limits_0^2 {{x^3}dx}\) using Simpson’s Rule with \(n = 4\) subintervals.

    Solution.

    The Simpson’s Rule formula for \(n = 4\) subintervals has the form

    \[{{S_4} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + 2\left( {{x_4}} \right)} \right].}\]

    Determine the width of the subinterval:

    \[{\Delta x = \frac{{b – a}}{n} }={ \frac{{2 – 0}}{4} }={ \frac{1}{2}.}\]

    The values of the cubic function at the points \({x_i} = \left\{ {0,\large{\frac{1}{2}}\normalsize,1,\large{\frac{3}{2}}\normalsize,2} \right\}\) are

    \[f\left( {{x_0}} \right) = f\left( 0 \right) = {0^3} = 0;\]

    \[f\left( {{x_1}} \right) = f\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{8};\]

    \[f\left( {{x_2}} \right) = f\left( 1 \right) = {1^3} = 1;\]

    \[f\left( {{x_3}} \right) = f\left( {\frac{3}{2}} \right) = {\left( {\frac{3}{2}} \right)^3} = \frac{{27}}{8};\]

    \[f\left( {{x_4}} \right) = f\left( 2 \right) = {2^3} = 8.\]

    Hence

    \[{\int\limits_0^2 {{x^3}dx} \approx {S_4} \text{ = }}\kern0pt{ \frac{1}{6}\left[ {0 + 4 \cdot \frac{1}{8} + 2 \cdot 1 + 4 \cdot \frac{{27}}{8} + 8} \right] }={ \frac{1}{6}\left[ {\frac{1}{2} + 2 + \frac{{27}}{2} + 8} \right] }={ \frac{{24}}{6} }={ 4.}\]

    Example 8.

    Using Simpson’s Rule with \(n = 4\) subintervals, evaluate the integral \(\int\limits_0^1 {{e^x}dx}.\) Round the answer to \(3\) decimal places.

    Solution.

    We evaluate the given integral by the formula

    \[{{S_4} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) }\right.}+{\left.{ 2\left( {{x_4}} \right)} \right].}\]

    Determine the width of the subinterval:

    \[\Delta x = \frac{{b – a}}{n} = \frac{{1 – 0}}{{4}} = \frac{1}{4}.\]

    Compute the function values at the endpoints of the subintervals:

    \[f\left( {{x_0}} \right) = f\left( 0 \right) = {e^0} = 1;\]

    \[{f\left( {{x_1}} \right) = f\left( {\frac{1}{4}} \right) = {e^{\frac{1}{4}}} = \sqrt[4]{e} }\approx{ 1.2840;}\]

    \[{f\left( {{x_2}} \right) = f\left( {\frac{1}{2}} \right) = {e^{\frac{1}{2}}} = \sqrt e }\approx{ 1.6487;}\]

    \[{f\left( {{x_3}} \right) = f\left( {\frac{3}{4}} \right) = {e^{\frac{3}{4}}} = \sqrt[4]{{{e^3}}} }\approx{ 2.1170;}\]

    \[f\left( {{x_4}} \right) = f\left( 1 \right) = {e^1} = e \approx 2.7183;\]

    Plugging in the function values into our equation, we get:

    \[{\int\limits_0^1 {{e^x}dx} \approx {S_4} }={ \frac{1}{{12}}\left[ {1 + 4 \times 1.2840 }\right.}+{\left.{ 2 \times 1.6487 }\right.}+{\left.{ 4 \times 2.1170 }\right.}+{\left.{ 2.7183} \right] }={ \frac{1}{{12}} \times 20.6197 }={ 1.7183 }\approx{ 1.718}\]

    Example 9.

    Using Simpson’s Rule with \(n = 2\) subintervals, evaluate the integral \(\int\limits_1^3 {\ln xdx}.\) Round the answer to \(3\) decimal places.

    Solution.

    We evaluate the integral by the following approximate formula:

    \[{{S_2} \text{ = }}\kern0pt{\frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + f\left( {{x_2}} \right)} \right].}\]

    The width of the subinterval is

    \[\Delta x = \frac{{b – a}}{n} = \frac{{2 – 0}}{{2}} = 1.\]

    Calculate the function values at the points \({x_i} = \left\{ {1,2,3} \right\}:\)

    \[f\left( {{x_0}} \right) = f\left( 1 \right) = \ln 1 = 0;\]

    \[f\left( {{x_1}} \right) = f\left( 2 \right) = \ln 2 \approx 0.6931;\]

    \[f\left( {{x_2}} \right) = f\left( 3 \right) = \ln 3 \approx 1.0986;\]

    Hence

    \[{\int\limits_1^3 {\ln xdx} \approx {S_2} }={ \frac{1}{3}\left[ {1 + 4\ln 2 + \ln 3} \right] }\approx{ \frac{1}{3}\left[ {1 + 4 \times 0.6931 + 1.0986} \right] }={ \frac{1}{3} \times 5.8710 }={ 1.957}\]

    Example 10.

    Using Simpson’s Rule with \(n = 4\) subintervals approximate the area under the sine curve \(f\left( x \right) = \sin x\) between \(x = 0\) and \(x = \pi.\)

    Solution.

    The Simpson’s Rule formula with \(n = 4\) segments is written in the form

    \[{{S_4} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) }\right.}+{\left.{ \left( {{x_4}} \right)} \right].}\]

    The width of the subinterval is

    \[{\Delta x = \frac{{b – a}}{n} }={ \frac{{\pi – 0}}{4} }={ \frac{\pi }{4}.}\]

    Calculate the values of the sine function at the points \({x_i} = \left\{ {0,\large{\frac{\pi }{4}}\normalsize, \large{\frac{\pi }{2}}\normalsize, \large{\frac{{3\pi }}{4}}\normalsize, \pi } \right\}:\)

    \[{f\left( {{x_0}} \right) = f\left( 0 \right) }={ \sin 0 }={ 0;}\]

    \[{f\left( {{x_1}} \right) = f\left( {\frac{\pi }{4}} \right) }={ \sin \frac{\pi }{4} }={ \frac{{\sqrt 2 }}{2};}\]

    \[{f\left( {{x_2}} \right) = f\left( {\frac{\pi }{2}} \right) }={ \sin \frac{\pi }{2} }={ 1;}\]

    \[{f\left( {{x_3}} \right) = f\left( {\frac{{3\pi }}{4}} \right) }={ \sin \frac{{3\pi }}{4} }={ \frac{{\sqrt 2 }}{2};}\]

    \[{f\left( {{x_4}} \right) = f\left( \pi \right) }={ \sin \pi }={ 0.}\]

    Now we can substitute these values into our equation and find the approximate value of the area:

    \[{A = \int\limits_0^\pi {\sin xdx} \approx {S_4} }={ \frac{\pi }{{12}}\left[ {0 + 4 \cdot \frac{{\sqrt 2 }}{2} + 2 \cdot 1 }\right.}+{\left.{ 4 \cdot \frac{{\sqrt 2 }}{2} + 0} \right] }={ \frac{\pi }{{12}}\left( {4\sqrt 2 + 2} \right) }={ \frac{\pi }{6}\left( {2\sqrt 2 + 1} \right).}\]

    Example 11.

    Using Simpson’s Rule with \(n = 4\) subintervals approximate the area under the inverse cosine curve \(f\left( x \right) = \arccos x\) between \(x = -1\) and \(x = 1.\)

    Solution.

    The Simpson’s Rule formula with \(n = 4\) segments has the form

    \[{{S_4} = \frac{{\Delta x}}{3}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) }\right.}+{\left.{ 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) }\right.}+{\left.{ \left( {{x_4}} \right)} \right].}\]

    Determine the width of the subinterval:

    \[{\Delta x = \frac{{b – a}}{n} }={ \frac{{1 – \left({-1}\right)}}{4} }={ \frac{1}{2}.}\]

    Calculate the values of the inverse cosine at the endpoints of the subintervals:

    \[{f\left( {{x_0}} \right) = f\left( { – 1} \right) }={ \arccos \left( { – 1} \right) }={ \pi ;}\]

    \[{f\left( {{x_1}} \right) = f\left( { – \frac{1}{2}} \right) }={ \arccos \left( { – \frac{1}{2}} \right) }={ \frac{{2\pi }}{3};}\]

    \[{f\left( {{x_2}} \right) = f\left( 0 \right) }={ \arccos 0 }={ \frac{\pi }{2};}\]

    \[{f\left( {{x_3}} \right) = f\left( {\frac{1}{2}} \right) }={ \arccos \frac{1}{2} }={ \frac{\pi }{3};}\]

    \[{f\left( {{x_4}} \right) = f\left( 1 \right) }={ \arccos 1 }={ 0.}\]

    Substituting these values into our equation, we find the area under the curve:

    \[{A = \int\limits_{ – 1}^1 {\arccos xdx} \approx {S_4} }={ \frac{1}{6}\left[ {\pi + 4 \cdot \frac{{2\pi }}{3} + 2 \cdot \frac{\pi }{2} }\right.}+{\left.{ 4 \cdot \frac{\pi }{3} + 0} \right] }={ \frac{1}{6} \cdot 6\pi }={ \pi }\]