GATE Papers >> Mechanical >> 2017 >> Question No 29

Question No. 29 Mechanical | GATE 2017

P (0,3), Q (0.5, 4), and R (1, 5) are three points on the curve defined by $\style{font-family:'Times New Roman'}{f(x)}$. Numerical integration is carried out  using both Trapezoidal rule and Simpson's rule within limits x=0 and x=1 for the curve. The difference between the two results will be

Answer : (A) 0

Solution of Question No 29 of GATE 2017 Mechanical Paper

Intration limits x E[0, 1]

No. of intervals = 2

$ \begin{array}{l}\mathrm h=\frac{1-0}2=0.5\\\mathrm P\left({\mathrm x}_0,\;{\mathrm y}_0\right)=\left(0.5,\;4\right)\\\mathrm Q\left({\mathrm x}_1,\;{\mathrm y}_1\right)=\left(0.5,\;4\right)\\\mathrm R\left({\mathrm x}_2,\;{\mathrm y}_2\right)=\left(1,\;5\right)\end{array} $

By Trapezoidal Method

$ \begin{array}{l}{\mathrm I}_1=\int\limits_0^1\mathrm f\left(\mathrm x\right)\operatorname d\mathrm x=\frac{\mathrm h}2\left(\left({\mathrm y}_0+{\mathrm y}_\mathrm n\right)+2\left({\mathrm y}_1+{\mathrm y}_2+...\right)\right)\\{\mathrm I}_1=\int\limits_0^1\mathrm f\left(\mathrm x\right)\operatorname d\mathrm x=\frac{\mathrm h}2\left[\left({\mathrm y}_0+{\mathrm y}_\mathrm n\right)+2\left({\mathrm y}_1\right)\right]\\{\mathrm I}_1=\frac{0.5}2\left[\left(3+5\right)+2\left(4\right)\right]\\{\mathrm I}_1=4\end{array} $

By simpson's$\frac13$ Rule

$ \begin{array}{l}{\mathrm I}_2=\int\limits_0^1\mathrm f(\mathrm x)\operatorname d\mathrm x=\frac{\mathrm h}3\left(\left({\mathrm y}_0+{\mathrm y}_\mathrm n\right)+4\left({\mathrm y}_1+{\mathrm y}_3+...\right)+2\left({\mathrm y}_2+{\mathrm y}_4+...\right)\right)\\{\mathrm I}_2=\frac{\mathrm h}3\left[\left({\mathrm y}_0+{\mathrm y}_2\right)+4\left({\mathrm y}_1\right)\right]\\{\mathrm I}_2=\frac{0.5}3(\left(3+5\right)+4\left(4\right))\end{array} $

Thus difference of results $ ={\mathrm I}_1-{\mathrm I}_2=4-4=0 $


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