Question-194548

Question Number 194548 by horsebrand11 last updated on 09/Jul/23

Commented by horsebrand11 last updated on 09/Jul/23

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Answered by AST last updated on 09/Jul/23

\frac{y - 1}{x - 1} = \frac{1 - 0}{1 - 0} \Rightarrow y = x (e q u a t i o n o f l i n e)

\int_{0}^{1} x d x = \frac{x^{2}}{2} ∣_{0}^{1} = \frac{1}{2}

x {(x - 2)}^{2} = x (x^{2} - 4 x + 4) = x^{3} - 4 x^{2} + 4 x

\Rightarrow \int_{1}^{2} (x^{3} - 4 x^{2} + 4 x) d x = \frac{x^{4}}{4} - \frac{4 x^{3}}{3} + 2 x^{2} ∣_{1}^{2} = \frac{5}{12}

\Rightarrow R_{2} = \frac{5}{12} + \frac{1}{2} = \frac{11}{12}

S i m i l a r l y, w e g e t \int_{0}^{2} x {(x - 2)}^{2} = \frac{4}{3} \Rightarrow R_{1} = \frac{4}{3} - \frac{11}{12} = \frac{5}{12}

Answered by MM42 last updated on 09/Jul/23

R_{1} = \int_{0}^{1} x {(x - 2)}^{2} d x - \frac{1}{2} = \frac{5}{12}

R_{2} = \int_{0}^{2} x {(x - 2)}^{2} d x - R_{1} = \frac{11}{12}

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