# 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)#

and

Please see below.

so the subintervals are

The corresponding midpoints are

Your values in the comments are correct:

So we need:

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

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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.

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.

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.

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