The correct option is B 16/3 I=∫_0^1x(1 x)^54dx #160; #10;—(1) #10; ∫_0^af(x)dx=∫_0^af(a x)dx #10; I=∫_0^11 xx^54dx=∫_0^1(x^ 54 x^ ...

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First Fundamental Theorem of Calculus

∫01x1-x5/4 dx...

Question

$\int_{0}^{1} \frac{x}{(1 - x)^{5 / 4}} d x =$

16 / 3

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3 / 16

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- 3 / 16

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- 16 / 3

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Solution

The correct option is B $- 16 / 3$
$I = \int_{0}^{1} \frac{x}{(1 - x)^{\frac{5}{4}}} d x$ ---(1)
$\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
$I = \int_{0}^{1} \frac{1 - x}{x^{\frac{5}{4}}} d x = \int_{0}^{1} (x^{\frac{- 5}{4}} - x^{\frac{- 1}{4}}) d x$
$= \int_{0}^{1} x^{\frac{5}{4}} d x - \int_{0}^{1} x^{\frac{- 1}{4}} d x$
$= \frac{(x^{\frac{- 5}{4} + 1})}{(\frac{- 5}{4} + 1)} \int_{0}^{1} - \frac{(x^{\frac{- 1}{4} + 1})}{(\frac{- 1}{4} + 1)} \int_{0}^{1}$
$= - 4 x^{\frac{- 1}{4}} \int_{0}^{1} - \frac{x^{\frac{3}{4}}}{\frac{4}{3}} \int_{0}^{1}$
$= - 4 - \frac{4}{3}$
$= \frac{- 16}{3}$

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\begin{matrix} {cos}^{4} \frac{π}{16} + {sin}^{4} \frac{π}{16} - {sin}^{2} \frac{π}{16} {cos}^{2} \frac{π}{16} + {cos}^{4} \frac{3 π}{16} + {sin}^{4} \frac{3 π}{16} - {sin}^{2} \frac{3 π}{16} {cos}^{2} \frac{3 π}{16} \end{matrix}

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