The area of the region above the x-axis bound | vedclass

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Question

(A) The given curve is $y = 	an x$ ... $(1)$When $x = \frac{\pi}{4}$,$y = 	an(\frac{\pi}{4}) = 1$. So,the point of tangency is $P(\frac{\pi}{4}, 1)$

Vedclass · Accepted Answer

$\frac{1}{2}\left( \log 2 - \frac{1}{2} \right)$

Vedclass · Answer

(A) The given curve is $y = 	an x$ ... $(1)$When $x = \frac{\pi}{4}$,$y = 	an(\frac{\pi}{4}) = 1$. So,the point of tangency is $P(\frac{\pi}{4}, 1)$.The derivative is $\frac{dy}{dx} = \sec^2 x$. At $x = \frac{\pi}{4}$,the slope $m = \sec^2(\frac{\pi}{4}) = 2$.The equation of the tangent at $P$ is $y - 1 = 2(x - \frac{\pi}{4})$,which simplifies to $y = 2x + 1 - \frac{\pi}{2}$ ... $(2)$.The $x-$intercept of the tangent is found by setting $y = 0$: $0 = 2x + 1 - \frac{\pi}{2} \implies x = \frac{\pi - 2}{4}$. Let this point be $L(\frac{\pi - 2}{4}, 0)$.The area of the region is the area under the curve $y = 	an x$ from $x = 0$ to $x = \frac{\pi}{4}$ minus the area of the triangle formed by the tangent line,the $x-$axis,and the vertical line $x = \frac{\pi}{4}$.Area $= \int_{0}^{\frac{\pi}{4}} 	an x \, dx - 	ext{Area of } \Delta PLM$$= [\log |\sec x|]_{0}^{\frac{\pi}{4}} - \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$= \log(\sec \frac{\pi}{4}) - \log(\sec 0) - \frac{1}{2} 	imes (\frac{\pi}{4} - \frac{\pi - 2}{4}) 	imes 1$$= \log(\sqrt{2}) - 0 - \frac{1}{2} 	imes (\frac{2}{4}) = \frac{1}{2} \log 2 - \frac{1}{4} = \frac{1}{2}(\log 2 - \frac{1}{2})$ sq units.

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

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