\(^9{∫}_1\) \(\frac{x(2x-3)}{√x}\) dx = \(\frac{2x^2 - 3x)}{x^{1/2}}\)
= x\(^{-1/2}\) (2x\(^2\) - 3x)
= 2x\(^{3/2}\) - 3x\(^{1/2}\)
= \(\frac{2x^{3/2}}{3/2 + 1}\) - \(\frac{3x^{1/2}}{1/2 + 1}\)
= \(\frac{2x^{3/2}}{5/2}\) - \(\frac{3x^{1/2}}{3/2}\)
= \(^9{∫}_1\) [\(\frac{4x^{3/2}}{5}\) - \(\frac{6x^{1/2}}{3}\)]
[ \(\frac{4(9)^{3/2}}{5}\) - \(\frac{6^(9){1/2}}{3}\) ] - [ \(\frac{4(9)^{1/2}}{5}\) - \(\frac{6(1)^{1/2}}{3}\) ]
= [ \(\frac{4(27)}{5}\) - \(\frac{6(3)}{3}\) - [ \(\frac{4}{5}\) - 2 ]
= [ \(\frac{108}{5}\) - 2 ] - [ \(\frac{6}{5}\) ]
= \(\frac{78}{5}\) + \(\frac{6}{5}\) = \(\frac{84}{5}\)
16\(^{4/5}\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}