`int_0^8sin(sqrt(x))dx, n = 4` Use the Midpoint Rule with the given value of `n` to approximate the integral. Round the answer to three decimal places.

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Chapter 5, 5.2 - Problem 9 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the midpoint rule to approximate the interval. First, you need to find `Delta x` , such that:

`Delta x = (b-a)/n`

The problem provides b=8, a=0 and n = 4, such that:

`Delta x = (8-0)/4 = 2`

Hence, the following intervals of length 2 are: `[0,2], [2,4], [4,6], [6,8].`

Now, you may evaluate the integral such that:

`int_0^8 sin (sqrt x) dx = Delta x(f((0+2)/2) + f((2+4)/2) + f((4+6)/2) + f((6+8)/2))`

`int_0^8 sin (sqrt x) dx = 2(f(1) + f(3) + f(5) + f(7))`

`int_0^8 sin (sqrt x) dx = 2(sin sqrt1 + sin sqrt3 + sin sqrt 5 + sin sqrt 7))`

`int_0^8 sin (sqrt x) dx = 2(0.017 + 0.030 + 0.039 + 0.046)`

`int_0^8 sin (sqrt x) dx = 2*0.132`

`int_0^8 sin (sqrt x) dx =0.264`

Hence, approximating the definite integral, using the midpoint rule, yields `int_0^8 sin (sqrt x) dx =0.264.`

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