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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œ‹/3]

PLEASE HELP ASAP

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Aravinda

  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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    oobleck

  2. i got 2.09, idk if it is right

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    Aravinda

  4. 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,
    ∫[0,Ο€/3] 4tan(x) dx = 2.77

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    oobleck

  5. 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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    oobleck

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