Posted on October 15, 2021October 24, 2022 by 202110150945 Exercises 8.1.A (Q1) For Exercises 1-6, find the area of the region bounded by the given curves. and Roughwork. are the points of intersection. The area of the shaded region is (alternatively)
and 
are the points of intersection.
of the shaded region is![\begin{aligned} A & = \int_{-1}^{3}|(2x+3)-(x^2)|\,\mathrm{d}x \\ & = \bigg[ x^2+3x-\frac{x^3}{3}\bigg]\bigg|_{-1}^{3} \\ & = \bigg( (3)^2+3(3)-\frac{(3)^3}{3}\bigg) - \bigg( (-1)^2+3(-1)-\frac{(-1)^3}{3}\bigg) \\ & = (9) - \bigg(-\frac{5}{3}\bigg) \\ & = \frac{32}{3}\quad \textrm{(squared units)} \\ \end{aligned}](https://physicspupil.com/wp-content/ql-cache/quicklatex.com-5f993d3e91bd421fd15165ddee343693_l3.svg "Rendered by QuickLaTeX.com")
![\begin{array}{cl} {A & = \displaystyle\int_{1}^{9}\bigg( \sqrt{y} - \frac{y-3}{2}\bigg)\,\mathrm{d}y + \int_{0}^{1}\Big( (\sqrt{y})-(-\sqrt{y})\Big)\,\mathrm{d}y }\\ & = \displaystyle\bigg( \frac{2y^{3/2}}{3}-\frac{y^2}{4}+\frac{3y}{2}\bigg)\bigg|_{1}^{9} + \bigg( \frac{4y^{3/2}}{3}\bigg)\bigg|_{0}^{1}} \\ & = \displaystyle\Bigg\{ \bigg[ \frac{2(9)^{3/2}}{3}-\frac{(9)^2}{4}+\frac{3(9)}{2}\bigg] -\bigg[\frac{2(1)^{3/2}}{3}-\frac{(1)}{4}+\frac{3(1)}{2}\bigg]\Bigg\} }\\ &\displaystyle\qquad \quad + \Bigg\{\bigg[\frac{4(1)^{3/2}}{3}\bigg] - \bigg[ \frac{4(0)^{3/2}}{3} \bigg]\Bigg\} }\\ & =\displaystyle \bigg( 18-\frac{81}{4}+\frac{27}{2}\bigg) - \bigg( \frac{2}{3}-\frac{1}{4}+\frac{3}{2} \bigg) + \bigg(\frac{4}{3}\bigg) - (0)} \\ & = \displaystyle\frac{32}{3}\quad \textrm{(squared units)}} \end{array}](https://physicspupil.com/wp-content/ql-cache/quicklatex.com-29c78f65521eca21cb00bbea7ceb330b_l3.svg "Rendered by QuickLaTeX.com")
