`int_1^5(x^2)(e^(-x))dx, n = 4` Use the Midpoint Rule with the given value of `n` to approximate the integral. Round the answer to four decimal places.

Textbook Question

Chapter 5, 5.2 - Problem 12 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the definite integral using the mid point rule, hence, first you need to evaluate `Delta x:`

`Delta x = (b-a)/n => Delta x = (5-1)/4 = 1`

You need to denote each of the 4 intervals, such that:` [1,2],[2,3],[3,4],[4,5].`

You need to evaluate the definite integral, such that:

`int_1^5x^*e^(-x)dx = Delta x*(f((1+2)/2)+f((2+3)/2)+f((3+4)/2)+f((4+5)/2))`

`int_1^5x^*e^(-x)dx = 1*(f(3/2) + f(5/2) + f(7/2) + f(9/2))`

`int_1^5x^*e^(-x)dx = ((3/2)^2*e^(-3/2) + (5/2)^2*e^(-5/2) + (7/2)^2*e^(-7/2) + (9/2)^2*e^(-9/2))`

`int_1^5x^*e^(-x)dx = (9/4*e^(-3/2) + 25/4*e^(-5/2) + 49/4*e^(-7/2) + 81/4*e^(-9/2))`

`int_1^5x^*e^(-x)dx = (2.25*0.21 + 6.25*0.08+ 12.25*0.03 + 20.25*0.01)`

`int_1^5x^*e^(-x)dx = (0.4725 + 0.5+ 0.3675 + 2.025)`

`int_1^5x^*e^(-x)dx =3.3650`

Hence, evaluating the definite integral, using the mid point rule, yields `int_1^5x^*e^(-x)dx =3.3650.`

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