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Question: If \(e^{0} = 1,\mspace{6mu} e^{1} = 2.72,\mspace{6mu} e^{2} = 7.39,\mspace{6mu} e^{3} = 20.09,\mspac...

If e0=1, 6mue1=2.72, 6mue2=7.39, 6mue3=20.09, 6mue4=54.60e^{0} = 1,\mspace{6mu} e^{1} = 2.72,\mspace{6mu} e^{2} = 7.39,\mspace{6mu} e^{3} = 20.09,\mspace{6mu} e^{4} = 54.60, then by Trapezoidal rule ∫04exdx=\int_{0}^{4}{e^{x}dx =}

A

53.87

B

53.60

C

58.00

D

None of these

Answer

58.00

Explanation

Solution

h=b−an=4−04=1h = \frac{b - a}{n} = \frac{4 - 0}{4} = 1

x :01234
y :12.727.3920.0954.60

By Trapezoidal rule

∫04f(x)dx=∫04exdx=h2[(y0+y4)+2(y1+y2+y3)]=12[(1+54.6)+2(2.72+7.39+20.09)]\int_{0}^{4}{f(x)dx = \int_{0}^{4}{e^{x}dx}} = \frac{h}{2}\left\lbrack (y_{0} + y_{4}) + 2(y_{1} + y_{2} + y_{3}) \right\rbrack = \frac{1}{2}\left\lbrack (1 + 54.6) + 2(2.72 + 7.39 + 20.09) \right\rbrack

=12[55.6+60.4]=58.00= \frac{1}{2}\lbrack 55.6 + 60.4\rbrack = 58.00