Question

ASAP. Please show all work thank you

The average (mean) value of ?(?) is given by

1/ ?−? ∫( b to a)?(?)??

on the interval (a,b) and is denoted by ????rage. Dont use another interval

(a) Find the average value of any quadratic function with vertex at the origin on the interval [?, ?].

(b) Find a number ? in the given interval such that ?(?) = ???erage

Answer 1

Key point here is, any quadratic with vertex at origin is of the form f(x)=kx^2

proof. Let f(x)= ax^2+bx+c, since f(0)=0 (it passes through origin) gives c=0. also slope at (0,0) is 0. f'(x)= 2ax+b putting f'(x) zero at zero gives b=0. hence f is of the form ax^2.

now the proof is easy

b qurage (mean value of f(x) is given by f f(x) dx on the internal [a, b] - ba a quadratic function with verlex at [0;o) looks like, both roots are zero. v КСо . k>0 =) f(x) is of the form kr². for some ke R1803 (9) f(x) = kx² b kn² dx faverage ) b-a b kx bra * cb² - 93) b-a J since since b>-4? =(b-a) (a+b79b) farrage = k. 62ab air bag 3 (b). Let for some 'c' f(C) = fang. kc? k. 976 ²7 ab -7 f(0) = a²+ b²+ ab Ca 27 => c ² a7b²tab 3. 3 It's to check CE[a,b] it's enough to show that ca & ccb. ca ccb => a3 26² abco 7 926²9b, a2 => (a-b) (a+2b) co =) which is also true since 7 Bab - 29? > 0 a-b<o. = (b-a)(b +2a)>0. which is pue since bab-a70 3