Let R be the region in the first quadrant that is enclosed by the graph of #y=tan x#, the x-axis, and the line x=pi/3, how do you find the area?

1 Answer
Nov 5, 2016

Answer:

ln2

Explanation:

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The area is given by integration with appropriate bounds

# :. A = int_0^(pi/3) tanx dx#
# :. A = int_0^(pi/3) sinx/cosx dx# .... [1]

Which we can integrate using a substitution

# { ("Let "u=cosx,=>,(du)/dx=-sinx), ("When "x=0,=>,u=cos0=1), ("And "x=pi/3,=>,u=cos(pi/3)=1/2) :} #

# :. int ... du = int ... -sinxdx #

Substituting into [1] we get

# A = int_1^(1/2) (-1/u)du #
# :. A = int_(1/2)^1 1/udu # (using #int_a^b...=-int_b^a...#)
# :. A = [lnu]_(1/2)^1 #
# :. A = ln(1)-ln(1/2) #

# :. A = 0-ln(1/2)_ # (as # ln(1) =0 #)
# :. A = ln(2) #