Practicing Success
The area of the region bounded by the curve $y =\frac{1}{1+(tanx)^{1/2}}$ and the x-axis between the ordinates x = π/6 and x = π/3 is |
π/4 π/2 π/8 none of these |
none of these |
The given curve is $y =\frac{1}{1+(tanx)^{1/2}}$ $⇒ y =\frac{(cosx)^{1/2}}{(sinx)^{1/2}+(cosx)^{1/2}}$ Area bounded between the intervals x = π/6 and x = π/3 is given by $A =\int\limits_{π/6}^{π/3}\frac{(cosx)^{1/2}}{(sinx)^{1/2}+(cosx)^{1/2}} dx$ $=\int\limits_{π/6}^{π/3}\frac{(cos(\frac{\pi}{3}+\frac{\pi}{6}-x))^{1/2}}{(sin(\frac{\pi}{3}+\frac{\pi}{6}-x))^{1/2}+(cos(\frac{\pi}{3}+\frac{\pi}{6}-x))^{1/2}}dx$ $=\int\limits_{π/6}^{π/3}\frac{(sinx)^{1/2}}{(sinx)^{1/2}+(cosx)^{1/2}} dx$ Adding $2A =\int\limits_{π/6}^{π/3}dx$ $⇒A = \frac{1}{2}[\frac{\pi}{3}-\frac{\pi}{6}]=\frac{\pi}{12}$ sq. units. Hence (D) is the correct answer. |