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Using a unit step size, the value of integral \(\mathop \smallint \limits_1^2 {\rm{x}}\ln {\rm{xdx}}\) by trapezoidal rule is ______

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 Concept:

Trapezoidal rule states that for a function y = f(x)

x

x0

x1

x2

x3

……

xn

y

y0

y1

y2

y3

……

yn

 

xn = x0 + nh, where n = Number of sub-intervals

h = step-size

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + {y_3} + \ldots + {y_{n - 1}}} \right)} \right]\)     …1)

For a trapezoidal rule, the number of sub-intervals must be a multiple of 1.

Calculation:

x = 1, y (0) = 1×ln1 = 0

x = 2, y (n) = 2ln2

x

1

2

y = x lnx

0

2ln2

\(\mathop \smallint \limits_1^2 {\rm{x}}\ln {\rm{xdx}} = \frac{1}{2}\left[ {0 + 2\ln 2} \right] = \ln 2 = 0.69\)
 

Key Points:

Apart from the trapezoidal rule, other numerical integration methods are;

A) Simpson’s one-third rule

B) Simpson’s three-eighth rule

A) Simpson’s one-third rule:

For applying this rule, the number of subintervals must be a multiple of 2.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_n}} \right) + 4\left( {{y_1} + {y_3} + {y_5} + \ldots + {y_{n - 1}}} \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots + {y_{n - 2}}} \right)} \right]\)     ..2)

B) Simpson’s three-eighths rule:

For applying this rule, the number of subintervals must be a multiple of 3.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{{3h}}{8}\left[ {\left( {{y_0} + {y_n}} \right) + 3\left( {{y_1} + {y_2} + {y_4} + {y_5} + \ldots } \right) + 2\left( {{y_3} + {y_6} + \ldots } \right)} \right]\)     …3)

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