# Calculus 2

Use the Midpoint Rule with n = 4 to approximate the area of the region bounded by the graph of the function and the x-axis over the given interval. (Round your answer to three decimal places.)
f(x) = 4 tan x, [0, π/3]

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1. so the interval is broken up at x = 0, Ο/12, Ο/6, Ο/4, Ο/3
each interval has width Ο/12, and each region is a trapezoid, with area
(f(x) + f(x+Ο/12))/2 * Ο/12 for x = 0, Ο/12, Ο/6, Ο/4
So review the midpoint rule (or trapezoid rule) to understand the shortcut for the computation.
Post your work if you get stuck

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2. i got 2.09, idk if it is right

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3. Sounds like you're stuck Why did you not show your work?

Actually, I got that wrong. The midpoint rule is not the same as the trapezoidal rule. We are still using rectangles. So the calculation is

Ο/12 (f(Ο/24)+f(3Ο/24)+f(5Ο/24)+f(7Ο/24)) = 2.74
Not sure how you got 2.09

Since the graph is concave up. a Riemann sum will underestimate. In fact,
β«[0,Ο/3] 4tan(x) dx = 2.77

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4. a handy calculator is at

www.emathhelp.net/en/calculators/calculus-2/midpoint-rule-calculator/?f=4tan%28x%29&a=0&b=pi%2F3&n=4

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