A=∫1211x-1dx x=cos2 θ⇒dx=-2sin θ cos θdθ x=1⇒θ=0, x=12⇒θ=π4 A=∫π40sec2θ-1(-2sin θ cos θ)dθ A=2∫0π4tan θsin θ cos θdθ A=∫0π42 sin2 θ dθ A=∫0π41- cos2 θ dθ A=θ+sin 2θ20π4=π4+12

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The area above the x-axis and under the curve $y = \sqrt{\frac{1}{x} - 1} for \frac{1}{2} \leq x \leq 1$ is

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$\frac{π}{4} - \frac{1}{2}$

$\frac{π}{4} + \frac{1}{2}$

$\frac{π}{4} + 1$

$\frac{π}{4} - 1$

answer is B.

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Detailed Solution

$A = \int_{\frac{1}{2}}^{1} \sqrt{\frac{1}{x} - 1} d x x = \cos^{2} θ \Rightarrow d x = - 2 \sin θ \cos θ d θ x = 1 \Rightarrow θ = 0, x = \frac{1}{2} \Rightarrow θ = \frac{π}{4} A = \int_{\frac{π}{4}}^{0} \sqrt{s e c^{2} θ - 1} (- 2 \sin θ \cos θ) d θ A = 2 \int_{0}^{\frac{π}{4}} \tan θ \sin θ \cos θ d θ A = \int_{0}^{\frac{π}{4}} 2 \sin^{2} θ d θ A = \int_{0}^{\frac{π}{4}} 1 - \cos 2 θ d θ A = {[θ + \frac{\sin 2 θ}{2}]}_{0}^{\frac{π}{4}} = \frac{π}{4} + \frac{1}{2}$