∫_1^2 √((x 1)(2 x))dx =∫_1^2 √((3x x^2 2))dx =∫_1^2 √((1/2)^2 (x 3/2)^2)dx =[(x 3/2)/2√((1/2)^2 (x 3/2)^2)+(1/2)^2/2sin^ 1((x 3/2)/1/2)]_1^2 =1/8( sin^ 11 s ...

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Byju's Answer

Standard XIII

Mathematics

Second Fundamental Theorem of Calculus

∫12 √x-12-xdx...

Question

$\int_{1}^{2} \sqrt{(x - 1) (2 - x)} d x =$

\frac{π}{8}

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\frac{π}{4}

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\frac{1}{8}

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none

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Solution

The correct option is A $\frac{π}{8}$
$\int_{1}^{2} \sqrt{(x - 1) (2 - x)} d x = \int_{1}^{2} \sqrt{(3 x - x^{2} - 2)} d x = \int_{1}^{2} \sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{3}{2})}^{2}} d x = {⎡ ⎣ \frac{(x - \frac{3}{2})}{2} \sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{3}{2})}^{2}} + \frac{{(\frac{1}{2})}^{2}}{2} {sin}^{- 1} (\frac{(x - \frac{3}{2})}{\frac{1}{2}}) ⎤ ⎦}_{1}^{- 1} = \frac{1}{8} ({sin}^{- 1} 1 - {sin}^{- 1} (- 1)) = \frac{π}{8}$

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∫21√(x−1)(2−x)dx=

The correct option is A π8∫21√(x−1)(2−x)dx=∫21√(3x−x2−2)dx=∫21√(12)2−(x−32)2dx=⎡⎣(x−32)2√(12)2−(x−32)2+(12)22sin−1((x−32)12)⎤⎦21=18(sin−11−sin−1(−1))=π8

$\int_{1}^{2} \sqrt{(x - 1) (2 - x)} d x =$