First slide

Area of bounded Regions

Question

$Find the area bounded by y = x^{3} - x and y = x^{2} + x$

Moderate

A

$37 s q u a r e u n i t s$
B

$\frac{37}{2} s q u a r e u n i t s$
C

$\frac{37}{3} s q u a r e u n i t s$
D

$\frac{37}{4} s q u a r e u n i t s$

Solution

$y = x^{3} - x = x (x - 1) (x + 1) is a cubic polynomial$ $function intersecting the x -axis at (- 1, 0), (0, 0), (1, 0) .$
$y = x^{2} + x = x (x + 1) is a quadratic function which is concave upward and intersect x -axis at (- 1, 0) (0, 0) .$
$The graphs of curves are as shown in following figure.$
$From the figure,$
$\begin{aligned} \begin{matrix} Requiredarea & = \int_{- 1}^{0} ((x^{3} - x) - (x^{2} + x)) d x + \int_{0}^{2} (x^{2} + x - (x^{3} - x)) d x \\ = \int_{- 1}^{0} (x^{3} - x^{2} - 2 x) d x + \int_{0}^{2} (x^{2} + 2 x - x^{3}) d x \\ = {[\frac{x^{4}}{4} - \frac{x^{3}}{3} - x^{2}]}_{- 1}^{0} + {[\frac{x^{3}}{3} + x^{2} - \frac{x^{4}}{4}]}_{0}^{2} \\ = [0 - (\frac{1}{4} + \frac{1}{3} - 1)] + [\frac{8}{3} + 4 - 4] \\ = \frac{37}{12} sq. units \end{matrix} \end{aligned}$

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