average value and substitution

Question

1.find the point at which the given function equal its average value on the given interval

f(x)=2-x^2; [-6,3]

2.use the substitution to evaluate the integral

(1)∫(dx)/√81-x^2 [0,1]

(2)∫(sinxdx)/(2+cosx)

Answer 1

average value of y= 2-x^2 over interval [-6,3]

find the definite integral, then divide by 9, length of the x interval

integral = 2x-x^3/3

evaluated from -6 to 3

2(3)-3^3/3 - [(2(-6)-(-6)^3/3]

= 6-9 -(-12+72)

= -3-60 =-57

divide by 9 to get -6 1/3= average y

-19/3 = 2 -x^2

x^2 = 2+19/3 = 25/3

x= +/-5/sqr3 = +/-5sqr3/3

points are (-5sqr3/3, -19/3) and (5sqr3/3, -19/3)

or a rough unreliable estimate

f(-6)= 2-36=-34= min

f(3)= 2-9=-7

f(0)=2 = max

f(1)=1

f(2)-2

f(-1)=1

f(-2)=2

f(-3)=-7

f(-4)=-14

f(-5)=-23

average y =about -8.7 f(-2.588)=-8.7 which isn't that far off from - 6.333...

about (-2.6, -8.7)