A certain breed of calf gains weight at a rate of 2.1√x pounds per week, where x is the age of the calf in weeks (0 ≤ x ≤ 16). Find the total weight gain of the calf between the ages of four weeks and nine weeks.
2007-05-14 09:18:09 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
integrate 2.1*x^.5 between the upper limit of 9 and lower limit of 4:
2.1*x^(3/2)/1.5 = 1.4*x^(3/2)
upper limit = 1.4*9^(3/2) = 37.8 lbs
lower limit =1.4*4^(3/2) = 11.2 lbs
total weight gain = 37.8-11.2 = 26.6 lbs
2007-05-14 09:27:17 · answer #1 · answered by minorchord2000 6 · 0⤊ 0⤋
f(x) = 2.1âx, pounds/week
the total weight gain of the calf between the ages of four weeks and nine weeks
= â«f(x) dx, x from 4 to 9
= 26.6 pounds
2007-05-14 16:39:52 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
Write down your data and your unknowns in a logical and well-organized way. Let's call the weight of the calf at week x the function w(x), since that's what we're studying. Thus, we are given the information dw/dx = 2.1Sqrt[x] and we want to know w(9) - w(4). We can get w(x) by integrating dw/dx w.r.t. x, where we get w(x) = (2/3)*2.1x^(3/2) + C where C is an unknown constant (actually the weight of the calf at week 0). Now you can find w(9) - w(4), since the C cancels itself out.
2007-05-14 16:26:16 · answer #3 · answered by Ron 6 · 0⤊ 0⤋
Let w(x) be the weight at any time x, with x measured in weeks. We're given a rate of 2.1âx, so dw/dx = 2.1âx. Integrate this to get an expression for w(x) in terms of x. Then find w(9) - w(4).
2007-05-14 16:26:52 · answer #4 · answered by Anonymous · 0⤊ 0⤋
W = 2.1 X^.5
I want to find the Integral of that equation from 4 to 9.
Integral = 2.1/1.5 * X^1.5 plus some constant. since we want gain only, we can ignore constant. Evaluate for 9 and 4 and subtract to get weight gain.
2007-05-14 16:26:51 · answer #5 · answered by Scott W 3 · 0⤊ 0⤋
Wooo! Let's do an integral!
You have the integral from 4 to 9 of 2.1âx dx.
The antiderivative of 2.1âx , or 2.1x^(1/2) is 1.4x^(3/2).
Then to evaluate the integral, we have
1.4*9^(3/2) - 1.4*4^(3/2) = 37.8 - 11.2 = 26.6 pounds.
2007-05-14 16:25:15 · answer #6 · answered by jesus.shaves 3 · 0⤊ 0⤋
Integral ( 2.1âx ) dx { between 4 and 9 }
2007-05-14 16:23:17 · answer #7 · answered by gjmb1960 7 · 0⤊ 0⤋
integrate 2.1*x^.5 between the upper limit of 9 and lower limit of 4:
2.1*x^(3/2)/1.5 = 1.4*x^(3/2)
upper limit = 1.4*9^(3/2) = 37.8 lbs
lower limit =1.4*4^(3/2) = 11.2 lbs
total weight gain = 37.8-11.2 = 26.6 lbs
2007-05-14 16:28:54 · answer #8 · answered by UrHero 3 · 0⤊ 0⤋