Question
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]
PLEASE HELP ASAP
f(x) = 4 tan x, [0, 𝜋/3]
PLEASE HELP ASAP
Answers
so the interval is broken up at x = 0, π/12, π/6, π/4, π/3
each interval has width π/12, and each region is a trapezoid, with area
(f(x) + f(x+π/12))/2 * π/12 for x = 0, π/12, π/6, π/4
So review the midpoint rule (or trapezoid rule) to understand the shortcut for the computation.
Post your work if you get stuck
each interval has width π/12, and each region is a trapezoid, with area
(f(x) + f(x+π/12))/2 * π/12 for x = 0, π/12, π/6, π/4
So review the midpoint rule (or trapezoid rule) to understand the shortcut for the computation.
Post your work if you get stuck
i got 2.09, idk if it is right
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
π/12 (f(π/24)+f(3π/24)+f(5π/24)+f(7π/24)) = 2.74
Not sure how you got 2.09
Since the graph is concave up. a Riemann sum will underestimate. In fact,
∫[0,π/3] 4tan(x) dx = 2.77
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
π/12 (f(π/24)+f(3π/24)+f(5π/24)+f(7π/24)) = 2.74
Not sure how you got 2.09
Since the graph is concave up. a Riemann sum will underestimate. In fact,
∫[0,π/3] 4tan(x) dx = 2.77
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
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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