Evaluate: \(^9{?}_1\) \(\frac{x(2x-3)}{?x}\) dx

Assessment: WAEC SSCE - Further Mathematics - 2021 (Essay) Subject: Further Mathematics

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Simplify the integrand first. Since \(\sqrt{x}=x^{1/2}\):

\[\frac{x(2x-3)}{\sqrt{x}}=\frac{2x^{2}-3x}{x^{1/2}}=2x^{3/2}-3x^{1/2}\]

Integrate:

\[\int_{1}^{9}\left(2x^{3/2}-3x^{1/2}\right)dx=\left[\frac{4}{5}x^{5/2}-2x^{3/2}\right]_{1}^{9}\]

At \(x=9\): \(9^{5/2}=3^{5}=243\) and \(9^{3/2}=27\), giving \(\tfrac45(243)-2(27)=194.4-54=140.4\).

At \(x=1\): \(\tfrac45-2=-1.2\).

\[140.4-(-1.2)=141.6=\frac{708}{5}\]

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