# calc: avg value

posted by .

Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2].

and this is what i did.. please check for mistakes. thanks :D

f(x) = x^2 sqrt(1+x^3), [0,2]
f ave = (1/(b-a))*inegral of a to b for: f(x) dx
f ave = (1/(2-0))*integral of 0 to 2 for: x^2 sqrt(1+x^3) dx
..let u = x^3 & du = 3x^2 dx
f ave = (1/2)*integral of 0 to 2 for: sqrt(1+u)*3 du
f ave = (3/2)*integral of 0 to 2 for: sqrt(1+u) du

= (3/2)[(2/3)(x+1)^(3/2)] from 0 to 2
= (3/2)[(2/3)((2)+1)^(3/2)] - (3/2)[(2/3)((0)+1)^(3/2)]
= (5.1962 - 1)
= 4.1962

* calc check: average value - bobpursley, Saturday, June 30, 2007 at 6:06am

When you change variables, you have to change limits of integration. When
x=0, u=0; when x=2, u=8

-------------------

How did you get u=8? wouldn't it be u=12 when x=2?

To find this don't I just plug in 2 for f(x), f(2)=(2)^2*Sqrt(1+(2)^2)) = 12
and then I would evaluate:

(3/2)[(2/3)(x+1)^(3/2)] at 0 and 12?

Please let me know if this is correct. Thanks.

NEVERMIND...dumb mistake on my part. I don't know what I was think. I know where the 0 and 8 come from. Thanks.

## Similar Questions

1. ### Calc, Mean Value Theorem

Consider the function : 3x^3 - 2x^2 - 4x + 1 Find the average slope of this function on the interval. By the Mean Value Theorem, we know there exists a "c" in the open interval (-2,3) such that f'(c) is equal to this mean slope. Find …
2. ### calc

Verify that the function satisfies the three hypotheses of Rolle's Therorem on the given interval. Then find all numbers c that satisfy the conclusiton of Rolle's Theorem. f(x)= x*sqrt(x+6) [-6,0] f is continuous and differential f(-6) …
3. ### calc check: average value

Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(b-a))*inegral of a to b for: f(x) dx f ave …
4. ### Calculus

Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 …
5. ### Math/Calculus

Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else?

Can someone check these for me? Please? Use half-angle identity to find the exact value of cos165 degrees. (-1/2) sqrt(2+sqrt(3)) Write the equation 2x+3y-5=0 in normal form. (-2sqrt(13)/13)x- (3sqrt(13)/13)y+ (5 sqrt (13)/13) = 0