For integral 2 15 e1/x dx. 1, the approximate value of T10 = 27.05 and M10 = 30.29
In this question, we need to find the approximations T10 and M10 for 2 15 e1/x dx. 1
i.e., To find: Integral approximate using Trapezoid rule and midpoint rule
f(x) = 15e^(1/x)
a = 1, b = 2, bx = 1/10
T10 = bx/2 [f(1) + 2f(1.1) + 2f(1.2) + 2f(1.3) + . . . + f(2) ]
= 1/20[15e + 30e^(1/1.1) + 30e^(1/1.2) + . . . + 15e^(1/2)]
= 1/20 * 30 [(e/2) + e^(1/1.1) + e^(1/1.2) + . . . + (e^(1/2) / 2)]
= 1.5 * 18.0362
= 27.05
Now we find M10
M10 = Δx [f(1.05) + f(1.15) + . . . + f(1.95)]
= 1/10 [15e^(1/1.05) + 15e^(1/1.15) + . . . + 15e^(1/1.95)]
= 15/10 [e^(1/1.05) + e^(1/1.15) + . . . + e^(1/1.95)]
= 1.5 * 20.1909
= 30.29
Therefore, for integral 2 15 e1/x dx. 1, the approximate value of T10 = 27.05 and M10 = 30.29
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