Question

for -25=co for osca Let I be the function defined by 1 (2) - {}(2+2) (2-2n The graph of fis shown in the figure above. Let R be the region bounded by the graph off and the ads (a) Find the area of R (b) Region R is the base of a solid. For this solid, each cross section perpendicular to the axis is a square. Write, but do not evaluate, an expression involving one or more integrals that gives the volume of the solid. axis to form a solid. Find (c) The portion of the region R for 1 Sys2 is revolved about the the volume of the solid

Let f be the function defined by F(x)=(1/2)(x+2)^2 for [-2,0) and 2-2sin(sqrtx) for [0, (x^3)/4]. the graph of f is shown in the figure above. Let R be the regiok bounded by the graph of f and the x-axis.

Answer 1

Let f be the function by feauna į ţ (x+21" for -2<x0 L 2-2 sinna ton osa - The graph of is shown it in the figure above let R be the legion hounded by the gaph of Demanis ☆ Find there area of R A and i Area = drea 1717 = fort die 1 = (set 25 ? dret (2-2sin Va] dz

Area *** 1{ (242) 97° +(22 * usin \e -40&ra) varying ? P-01 + (8 02 +45 - I - tes=) 13) - cotesino – 4 C0567 0) 3 + + 4 N 6 Regim R Regim R is base of solid for this solid each closs sed section Pexpendicular to the l-axis is a square to each couss sectius perpendicular to the T axis is a square: Carea of cliss sedini) : Volume = jº te j * dx + ] [ 2.25in ca] de [Velone = 9 (42)" dve + 48*(1 – 2 sin v tusirining 274 4.

The volume of the said ... formed by evolving the regim about x-axis is 190]{+ C70x13?) Loc it U VETT dx -2 V2 vr -ci). ${{ esinej: orben] ** 1 (9*_ ) 14+ J'te cartongest 2 + - 16 Sin VI - 16 COS(VT) Cos(265) +2Sin (2724).V-22 | 4 <2012 - +46 Los == [ 4og 2-ve] LS-10 in 5 – 1603) , - bus ms + 2 sin (m). 17-2022] 20+ 2-1 - = + (x2 +2= 1 [ -1]+(56-164 - 8 7.96-4-2 3.76 at +2 -12 [- 1)]+**** is tots} IV - [ the ] = + (8 + + + 3y3m]