The area of the region (in square units) above the x-axis bounded by the curve $y=\tan x,0\le x\le \frac{\pi }{2}$  and the tangent to the curve at $x=\frac{\pi }{4}$ is  
 

Required area $=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(\tan x\right)dx$
-(Area of triangle formed by tangent $x=\frac{\pi }{4}andx-axis$,  )
 $$\begin{gathered} =[\log \left(\sec \frac{\pi }{4}\right)-\log \left(\sec 0\right)]-\frac{1}{2} \times 1(\frac{\pi }{4}-x_0) \\ =\log \sqrt{2}-\frac{1}{4}(where{x}_0=\frac{\pi }{4}-\frac{1}{2}isx-\text{intercept\hspace{0.17em}of\hspace{0.17em}tangent}) \\ =\frac{1}{2}\log 2-\frac{1}{4} \end{gathered}$$