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
![\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")
(alternatively)
![\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")
物理子衿
物理子衿
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)