# 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

ln2

#### Explanation:

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

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) #