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value of the definite integral \int_{2}^{3} [ \sqrt{ 2x - \sqrt{5 ( 4x -5 )}}+ \sqrt{ 2x + \sqrt{5 ( 4x -5 )}}] dx is equal to

Option: 1

4\sqrt 3 - 2 \sqrt 2 /3


Option: 2

4 \sqrt 2


Option: 3

4 \sqrt 3 - 4/3


Option: 4

\frac{\sqrt 10 }{2} + \frac{7 \sqrt 7 - 5 \sqrt 5 }{3 \sqrt 2 }


Answers (1)

best_answer

 

Indefinite integrals for Algebraic functions -

 \frac{\mathrm{d}}{\mathrm{d} x} \frac{\left ( x^{n+1} \right )}{n+1}=x^{n} so \int x^{n}dx=\frac{x^{n+1}}{n+1}

- wherein

Where  n\neq-1

 

 

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

 

 

- wherein

Since \int f(x)dx=\int f(t)dt=\int f(\theta )d\theta all variables must be converted into single variable ,\left ( t\, or\ \theta \right )

 

 

 

Let \: 5(4x -5) =t^{2}          \Rightarrow           20x - 25 = t^2 

  x =t^2 + 25 / 20

            Also\: : 20 dx = 2t dt \: or\: dx =t/10 dt

               \therefore I = \int_{\sqrt {15}}^{\sqrt {35}}\left ( \sqrt{\frac{t^2 +25 }{10}}- t + \sqrt{\frac{t^2 + 25 }{10 }}+t \right ) \frac{t}{10} dt

\int_{\sqrt {15}}^{\sqrt {35}}\left ( \frac{|t-5|}{\sqrt {10}} + \frac{|t+5 |}{\sqrt {10}}\right ) \frac{t}{10}dt \\\\

\\=\int_{\sqrt {15}}^{{5}}\left ( \frac{-t+5}{\sqrt {10}} + \frac{t+5}{\sqrt {10}}\right ) \frac{t}{10} dt +\int_{5}^{\sqrt{35}}\left ( \frac{t-5}{\sqrt {10}} + \frac{t+5}{\sqrt {10}}\right ) \frac{t}{10} dt\\\Rightarrow \frac{1}{\sqrt{10}}\left [ \frac{t^{2}}{2} \right ]^{5}_{\sqrt{15}}+\left[\frac{t^3}{15\sqrt{10}}\right]_{_5}^{^{\sqrt{35}}}\\

 

 ??????=\frac{1}{\sqrt10}\left [ \frac{25}{2}-\frac{15}{2} \right ]+\frac{1}{15\sqrt{10}}\left [ \left ( \sqrt{35} \right )^{3}-125 \right ]

=\frac{\sqrt{10}}{2}+\frac{7\sqrt{7}-5\sqrt{5}}{3\sqrt{2}}

 

              

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manish

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