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Approximate #\int_0^2(1)/(1+x^3)dx# using the Midpoint Rule, given #n=4#? Please check my work

#f(\barx_1)=3/4#
#f(\barx_2)=11/36#
#f(\barx_3)=37/504#
#f(\barx_4)=93/3640#

and #\Deltax=(b-a)/n=(2-0)/4=1/2#

#\thereforeM_4=1/2(3/4+11/36+37/504+93/3640)#

Answer 1

Christopher Agresta

Please see below.

On interval #[0,2]# with #n = 4#, we get #Deltax = (2-0)/4 = 1/2#

so the subintervals are

#[0,1/2]#, #[1/2,1]#, #[1,3/2]#, #[3/2,2]#

The corresponding midpoints are

#1/4#, #3/4#, #5/4# and #7/4#.

each of these has the form #a/4# (#a = 1, 3, 5, 7#)

#f((a/4)) = 1/(1+(a/4)^3) = 1/(1+(a^3/64)) = 1/((64+a^3)/64)#

#f(a/4) = 64/(a^3+64)#.

Your values in the comments are correct:

So we need:

#(64/65+64/91+64/189+64/407) * 1/2 = (32/65+32/91+32/189+32/407)#

#65 = 5 * 13# #91 = 7 * 13# #189 = 9 * 21 = 7 * 27# #407 = 11 * 37#

The least common factor of the denominators is: #5 * 7 * 11 * 13 * 27 * 37#

Now grind out the arithmetic. (And be happy you were born in the age of electronic calculators. Math students used to have to do this stuff all the time.)

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Answer 2

Christopher Agresta

Using the Midpoint Rule with ( n = 4 ), the integral approximation is approximately ( 1.2674 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Approximate #\int_0^2(1)/(1+x^3)dx# using the Midpoint Rule, given #n=4#? Please check my work

#f(\barx_1)=3/4# #f(\barx_2)=11/36# #f(\barx_3)=37/504# #f(\barx_4)=93/3640# and #\Deltax=(b-a)/n=(2-0)/4=1/2# #\thereforeM_4=1/2(3/4+11/36+37/504+93/3640)#

#f(\barx_1)=3/4#
#f(\barx_2)=11/36#
#f(\barx_3)=37/504#
#f(\barx_4)=93/3640#

and #\Deltax=(b-a)/n=(2-0)/4=1/2#

#\thereforeM_4=1/2(3/4+11/36+37/504+93/3640)#