The midpoint rule can be stated as below.

if n - number of intervals, the interval is , `Deltax`

`Deltax = (b-a)/n`

where, b - upper limit of integral and and a is the lower limit of integral.

then if for each interval, `x_1, x_2,....x_n` , the midpoint be `(x_1^*,x_2^*...x_n^*)`

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The midpoint rule can be stated as below.

if n - number of intervals, the interval is , `Deltax`

`Deltax = (b-a)/n`

where, b - upper limit of integral and and a is the lower limit of integral.

then if for each interval, `x_1, x_2,....x_n` , the midpoint be `(x_1^*,x_2^*...x_n^*)`

`int_a^bf(x)dx = Deltax(f(x_1^*)+f(x_2^*)+....+f(x_n^*))`

now for our, problem, (b-a) is (pi/4 -0).

then, `Deltax = (pi/4)/4 = pi/16`

therefore the intervals are, `(0,pi/16), (pi/16,pi/8) and (pi/8,pi/4)`

the midpoints are,

`pi/32, (3pi)/32 and (5pi)/32`

therefore the midpoint values,

`2cos^5(pi/32) = 1.952309`

`2cos^5((3pi)/32) = 1.604923`

`2cos^5((5pi)/32) = 1.067035`

according to definiton,

`int_0^(pi/4)2cos^5(x)dx = (pi/16)(1.952309+1.604923+1.067035))`

`int_0^(pi/4)2cos^5(x)dx = 0.9079`

as accurate to 4 decimal places, (The actual anser is 1.01352, so this is an approximation)