Question

4. (Calculator) Let R be the region bounded by the graphs of f(x)= 20+x-x2 and g(x)=x-5x. (a) Find the area of R. (b) A vertical line x k divides R into two regions of equal area. Write, but do not solve, an equation that could be solved to find the value of k (c) The region R is the base of a solid. For this solid, the cross sections perpendicular to the x-axis are isosceles right triangles with the hypotenuse as the base of the solid. Write, but do not evaluate, an integral expression for the volume of this solid. (d) Write, but do not evaluate, an about the horizontal line y-10 (e) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotated about the vertical line x- 7. Graph for Problem 4 S integral expression for the volume of the solid generated when R is rotated

Answer 1

4) f(*) = 20 x- n 2 aınd a) g) f)=g() 2) 20+x-v2 = x2-sx 2 3x-0 = O -) S Area -2x7dr 2 o 6Y 2 -2 20 x 3x2 - 2x3 34 3 3 -2 h haf b) Cerea 2 =K une dvcden It()-gd 343 - 2 343 3 3 24 -20x -2 -2K3 +3k<20k + 62 3 34-3 2 K3 3 K2-2 OK - 69 2 2 hypotenuu 4- Le - c) Area ot isoce lis vightriang 20+6xe 2x2 20+6x-2z2 f(x)-l) Hy poteruu Asca o Aerpındiwlar ring e toz - axi o+ 6z - 2x 2) (103x -x2)4 4- Cio +3x-2 dx Voume -2

ニー10 OuterRacius fix) 0) Innur Racuu x2-Sx to 304-2 2-Sx+ 107 dx Voune 2. S (x2-x-30)-x2- Sx+10)dx/ -2 e) 7- x Radiu Hugt ()-( mH Radius 5 2K (7-x)L20 +6x- 2x2dlk. Vo ume