Let α>0. If ∫0α xx+α−xdx=16+20215, then α is equal to :

∫0α xx+α−xdx=16+202151α∫0αx (x+α+x)dx1α⋅∫0α (x+α−α)x+α+x321α⋅∫0α (x+α)32−α(x+α)12+x32dx=1α⋅(x+α)52⋅25−α⋅23(x+α)32+x52⋅250α1α25(2α)5/2−2α3(2α)3/2+25α5/2−25α5/2+23α5/2=1α27/2α5/2525/2α5/23+23α5/2=α3/227/25−25/23+23==α3/215(242−202+10)=α3/215(42+10)=αα(10+42)15=16+20215⇒α=2α=2

Let α>0. If ∫0α xx+α−xdx=16+20215, then α is equal to :

∫0α xx+α−xdx=16+202151α∫0αx (x+α+x)dx1α⋅∫0α (x+α−α)x+α+x321α⋅∫0α (x+α)32−α(x+α)12+x32dx=1α⋅(x+α)52⋅25−α⋅23(x+α)32+x52⋅250α1α25(2α)5/2−2α3(2α)3/2+25α5/2−25α5/2+23α5/2=1α27/2α5/2525/2α5/23+23α5/2=α3/227/25−25/23+23==α3/215(242−202+10)=α3/215(42+10)=αα(10+42)15=16+20215⇒α=2α=2

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Let $α > 0 . If \int_{0}^{α} \frac{x}{\sqrt{x + α} - \sqrt{x}} dx = \frac{16 + 20 \sqrt{2}}{15}, then α$ is equal to :

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$2 \sqrt{2}$

$\sqrt{2}$

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Detailed Solution

$\begin{array}{l} \int_{0}^{α} \frac{x}{\sqrt{x + α} - \sqrt{x}} dx = \frac{16 + 20 \sqrt{2}}{15} \\ \frac{1}{α} \int_{0}^{α} x (\sqrt{x + α} + \sqrt{x}) dx \\ \frac{1}{α} \cdot \int_{0}^{α} ((x + α - α) \sqrt{x + α} + x^{\frac{3}{2}}) \\ \frac{1}{α} \cdot \int_{0}^{α} ((x + α)^{\frac{3}{2}} - α (x + α)^{\frac{1}{2}} + x^{\frac{3}{2}}) dx \\ = \frac{1}{α} \cdot {((x + α)^{\frac{5}{2}} \cdot \frac{2}{5} - α \cdot \frac{2}{3} (x + α)^{\frac{3}{2}} + x^{\frac{5}{2}} \cdot \frac{2}{5})}_{0}^{α} \\ \frac{1}{α} (\frac{2}{5} (2 α)^{5 / 2} - \frac{2 α}{3} (2 α)^{3 / 2} + \frac{2}{5} α^{5 / 2} - \frac{2}{5} α^{5 / 2} + \frac{2}{3} α^{5 / 2}) \\ = \frac{1}{α} (\frac{2^{7 / 2} α^{5 / 2}}{5} \frac{2^{5 / 2} α^{5 / 2}}{3} + \frac{2}{3} α^{5 / 2}) \\ = α^{3 / 2} (\frac{2^{7 / 2}}{5} - \frac{2^{5 / 2}}{3} + \frac{2}{3}) \\ = = \frac{α^{3 / 2}}{15} (24 \sqrt{2} - 20 \sqrt{2} + 10) = \frac{α^{3 / 2}}{15} (4 \sqrt{2} + 10) \\ = α \sqrt{α} \frac{(10 + 4 \sqrt{2})}{15} = \frac{16 + 20 \sqrt{2}}{15} \Rightarrow α = 2 \\ α = 2 \end{array}$