First slide

Area of bounded Regions

Question

$The area bounded by the curves y = x^{2} - 2 x + 2; x \geq 1 and its inverse is given by$

Moderate

A

$\frac{1}{3}$
B

$\frac{2}{3}$
C

$\frac{5}{3}$
D

$\frac{4}{3}$

Solution

$y = {(x - 1)}^{2} + 1$
$\begin{array}{l} P u t y = x \\ x = x^{2} - 2 x + 2 \\ x^{2} - 3 x + 2 = 0 \\ (x - 1) (x - 2) = 0 \\ x = 1, 2 \\ Re q u i r e d a r e a = 2 \int_{1}^{2} (x - {(x - 1)}^{2} - 1) d x \\ = 2 {[\frac{x^{2}}{2} - \frac{{(x - 1)}^{3}}{3} - x]}_{1}^{2} \\ = 2 [2 - \frac{1}{3} - 2 - \frac{1}{2} + 1] \\ = 2 [1 - \frac{5}{6}] = 2 (\frac{1}{6}) = \frac{1}{3} \end{array}$

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