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By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

    Q9.    \int_0^2x\sqrt{2-x}dx

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We have                 I\ =\ \int_0^2x\sqrt{2-x}dx

By using the identity  

                                          \ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get : 

                                     I\ =\ \int_0^2x\sqrt{2-x}dx\ =\ \int_0^2(2-x)\sqrt{2-(2-x)}dx

or                                 I\ =\ \int_0^2(2-x)\sqrt{x}dx

or                                 I\ =\ \int_0^2(2\sqrt{x}\ -\ x^\frac{3}{2} dx

or                                       =\ \left [ \frac{4}{3}x^\frac{3}{2}\ -\ \frac{2}{5}x^\frac{5}{2} \right ]^2_0

or                                      =\ \frac{4}{3}(2)^\frac{3}{2}\ -\ \frac{2}{5}(2)^\frac{5}{2}

or                               I\ =\ \frac{16\sqrt{2}}{15}

Posted by

Devendra Khairwa

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