The estimate of \(\int_{0.5}^{1.5}\dfrac{dx}{x}\) obtained using Simpson's rule with three-point function evaluation exceeds the exact value by

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  1. 0.235
  2. 0.068
  3. 0.024
  4. 0.012

Answer (Detailed Solution Below)

Option 4 : 0.012
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Concept:

The Simpson’s rule is given by,

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

h –Width of interval / Step length

y0, y1, …yn – Ordinates corresponding to x0, x1, ……xn

Error =|Exact value – Approximate value|

Calculation:

Given:

\({\rm{I}} = \mathop \smallint \nolimits_{0.5}^{1.5} \dfrac{{{\rm{dx}}}}{x}\)

Three point function ∴ 0.5, 1, 1.5

h = 0.5

x = 0.5, y = 2

x = 1, y = 1

x = 1.5, y = 0.666

x

0.5

1

1.5

y

2

1

0.666

 

\(\mathop \smallint \nolimits_{0.5}^{1.5} \dfrac{{dx}}{x} = \dfrac{{0.5}}{3}\left[ {\left( {2 + 0.666} \right) + 4\left( 1 \right)} \right] = 1.111\)

The exact value of the given integral is

\(\mathop \smallint \nolimits_{0.5}^{1.5} \dfrac{{dx}}{x} = \left[ {\ln x} \right]_{0.5}^{1.5} = \ln 1.5 - \ln0.5 = 1.0986\)

Exact value = 1.0986

Error =|1.0986 –1.111| = 0.012
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