Find the area of the region bounded by C: y = tanx , tangent drawn at C at x = pie/4 and the x- axis.

Answer 1

We will need the equation of the tangent.
y = tanx
y' = sec^2 x
when x = π/4 , tan π/4 = 1
so at the point (π/4,1)
the slope is sec^2 (π/4) = 2
so equation is y = 2x + b
at (π/4,1)
1 = 2(π/4) + b
b = 1 - π/2 = (2-π)/2

tangent equation is y = 2x + (2-π)/2
the x-intercept, let y = 0
2x = (π-2)/2 , x = (π-2)/4

I hope you made a sketch
I am going to take the area below y = tanx from 0 to π/4, then subtract the little right-angled triangle.

first part of area
= ∫tanx dx from 0 to π/4
= -ln(cosx) | from 0 to π/4
= (-ln(1/√2) - (-ln(1) )
= - ( ln1- ln√2) + 0
= ln √2

base of little triangle = π/4 - (π-2)/4
= (π - π + 2)/4 = 1/2
height = 1
area = (1/2(1/2)(1) = 1/4

area you want =ln(√2) - 1/4 or
= (4ln√2 - 1)/4
= 2ln2 - 1)/4 or
= (ln4 - 1)/4