Finding average of intervals

Question

My daughter has worked this problem and thinks she has the answer figured out but the online assignment isn't accepting her answer as being correct. Can someone please verify her work and point out if/where she strayed off? The problem is:

Find the number(s) b such that the average value of f(x)=6x^2-42x+52 on the interval (0,b) is equal to 18.

She came up with the following solution using the quadratic formula:

18=6x^2-42x+52

0=6x^2-42x+34

(-(-42)(+/-)sqr((-42)+4(6)(34)))/(6(2))

(42(+/-)sqr(1764+816))/12

(42=sqr2380)/12 and (42-sqr2380)/12

0.934199, 6.0658

Answer 1

Need to use the average over an interval integral. Set that to 18 and solve for b.

Avg = ∫^b_af(x)dx /(b-a)

∫^b₀6x²-42x+52 dx /(b-0) = 2b²-21b+52 = 18

2b²-21b+34=0 ==> b = 8.5, 2

Answer 2

Hi Steven, show this to your daughter:

Avg value= (1/(b-a)) ∫₀^b f(x)dx where f(x)=6x²-42x+52, a=0, and Avg value=18

18=1/b ∫₀^b 6x²-42x+52 dx

Reverse power rule: 18=1/b( 2x³ -21x² +52x |₀^b ) ------------->(2b³ -21b² +52b)/b=18

2b² -21b+52=18. Subtract 18 from both sides: 2b^2-21b+34=0. Factor by splitting the middle term:

(2b-17)(b-2)=0 --------->b=8.5 and b=2

I hope this helps her!