English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Please explain to me whether the statement below is true or false.
1. On the interval [a,b], the average value of f(x) + g(x) is the average
value of f(x) plus the average value of g(x).
2. If a than the average value of f on the interval [a,b]
3. The average value of the product, f(x)g(x), of two functions on
an interval equals the product of the average values of f(x) and
g(x) on the interval.
4. A 4-term left hand Riemann sum approximation cannot give the
exact value of a definite integral.

Many thanks to those who are willing to help me.

2006-11-08 08:41:48 · 4 answers · asked by ? 1 in Science & Mathematics Mathematics

4 answers

1)true
2)false
3)false
4)false

2006-11-08 08:53:58 · answer #1 · answered by 11:11 3 · 0 0

1. Write equation for average of f+g and see if it separates into the sum of the averages (it does)
2. I don't think so. Cook up an extreme function to see:
f= 0 for [a,c] and [d,b] and f=10000 for [c,d] and make a = -100, b = -1, c=+1, d = +100

You do the arithmetic.

2006-11-08 08:50:32 · answer #2 · answered by modulo_function 7 · 0 0

Trig substitution is unquestionably mandatory good right here. you %. x^2 to equivalent (9/sixteen)(tan?)^2 so which you would be waiting to simplify the denominator, so enable... x = (3/4)tan ? dx = (3/4)(sec ?)^2 d? Now combine... ?[(3/4)(sec ?)^2]d?/[9 + sixteen(9/sixteen)(tan ?)^2] (3/4)?(sec ?)^2 d?/[9(a million + (tan?)^2)] (3/36)?(sec ?)^2 d?/[(sec?)^2] (a million/12)?d? (a million/12)? + C submit to in recommendations that... x = (3/4)tan? So... tan? = (4x/3) ? = arctan(4x/3) So the fundamental is unquestionably... (a million/12)arctan(4x/3) +C with any success you spot the situation you went incorrect.

2016-10-15 13:13:03 · answer #3 · answered by ? 4 · 0 0

1.true
2.false
3.false
4.false

2006-11-08 09:26:00 · answer #4 · answered by maggieissohot 1 · 1 0

fedest.com, questions and answers