# 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?

ln2

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) #

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To find the area of the region ( R ) enclosed by the graph of ( y = \tan(x) ), the x-axis, and the line ( x = \frac{\pi}{3} ) in the first quadrant, you need to integrate the function ( \tan(x) ) over the interval ([0, \frac{\pi}{3}]) and then take the absolute value of the result.

The integral to find the area ( A ) is given by:

[ A = \int_{0}^{\frac{\pi}{3}} |\tan(x)| , dx ]

Since ( \tan(x) ) is positive in the first quadrant, taking the absolute value is not necessary. We can directly integrate ( \tan(x) ) from ( 0 ) to ( \frac{\pi}{3} ):

[ A = \int_{0}^{\frac{\pi}{3}} \tan(x) , dx ]

You can then evaluate this integral to find the area of the region ( R ).

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