True or False Integration?

Question

I need an explanation of the statement below if it is true or false:

If f(x) < g(x) on the interval [a,b], then the average value of f is less than or equal to the average value of g on the interval [a,b]

Answer 1

This is true additionally it is less(cannot be equal)

reason is f(x) = g(x)

ingegartion is arera under curve = sum f(x)(h etc

area under curve for each range is less for f than g

so avarage of f < avarage of g

Answer 2

Proving the obvious, eh? Not always easy.
Does average value here mean the integral of the function divided by the length of the function? If so, you need only to prove that the integral of f is less than the integral of g.
Now intl(f) - intl(g) = intl(f - g), and since f - g < 0 for all x in the interval, we've shown that
intl(f) - intl(g) < 0
i.e. intl(f) < intl(g)

Answer 3

if f(x) f(c) {f(c1)+f(c2)+f(c3)....f(cn)}/n
since f(c1)

Answer 4

True.

The average value of f = 1/(b-a) * integral from a to b of f(x)

The average value of g = 1/(b-a) * integral from a to b of g(x)

The integral from a to b of f(x) is less than the integral from a to b of g(x) so the average value of f on (a,b) is less than the average value of g on (a,b)

Answer 5

one million. Write equation for popular of f+g and notice if it separates into the sum of the averages (it does) 2. i don't think of so. cook dinner up an severe functionality to work out: f= 0 for [a,c] and [d,b] and f=10000 for [c,d] and make a = -a hundred, b = -one million, c=+one million, d = +a hundred You do the mathematics.

Answer 6

it is true. look up mean value theorem.

the mean value of f(x) will be some point c such that f(c) exists. Also, the mean value of g(x) will be at some point k such that g(k) exists. since we already know f(x) is less then g(x) for all x in [a,b] then it must be that f(c) , the mean of f(x), is less then g(k), the mean of g(x).

Answer 7

HY was right