Help Pa sagot po pls
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We can start by finding the definite integral of csc^2 x over the given interval, then divide the result by the length of the interval to get the average value.
Using a trigonometric identity, csc^2 x = 1 + cot^2 x, we can rewrite the integrand as:
1 + cot^2 x
The antiderivative of cot^2 x can be found using substitution:
Let u = cot x, then du/dx = -csc^2 x dx, so dx = -du/(u^2+1)
Substituting and integrating, we get:
∫cot^2 x dx = -∫du/(u^2+1) = -arctan u = -arctan(cot x)
Going back to the original integral, we have:
∫(1+cot^2 x) dx = x - arctan(cot x) + C
where C is the constant of integration.
Evaluating this expression at the limits of integration, we get:
[π/4 - arctan(1)] - [π/б - arctan(б)] = π/4 - π/б + arctan(б) - arctan(1)
Using the formula arctan(a) - arctan(b) = arctan((a-b)/(1+ab)), we can simplify this to:
π/4 - π/б + arctan(б) - arctan(1) = π/4 - π/б + arctan(б-1/(1+б))
Now we can find the length of the interval by subtracting the endpoints:
π/4 - π/б = (π/4б)(4-б)
Therefore, the average value of csc^2 x over the given interval is:
[π/4 - π/б + arctan(б-1/(1+б))] / [(π/4б)(4-б)]
To determine which of the given choices is the closest to this value, we need to plug in values of б and compare the results. We can start by testing the middle value, б = 2:
[π/4 - π/2 + arctan(3/5)] / [(π/8)(2)] ≈ 2.796
This is closest to choice (C), so our final answer is:
Average of csc^2 x over the interval from x = π/б tо x = π/4 is approximately 2.796.
Using a trigonometric identity, csc^2 x = 1 + cot^2 x, we can rewrite the integrand as:
1 + cot^2 x
The antiderivative of cot^2 x can be found using substitution:
Let u = cot x, then du/dx = -csc^2 x dx, so dx = -du/(u^2+1)
Substituting and integrating, we get:
∫cot^2 x dx = -∫du/(u^2+1) = -arctan u = -arctan(cot x)
Going back to the original integral, we have:
∫(1+cot^2 x) dx = x - arctan(cot x) + C
where C is the constant of integration.
Evaluating this expression at the limits of integration, we get:
[π/4 - arctan(1)] - [π/б - arctan(б)] = π/4 - π/б + arctan(б) - arctan(1)
Using the formula arctan(a) - arctan(b) = arctan((a-b)/(1+ab)), we can simplify this to:
π/4 - π/б + arctan(б) - arctan(1) = π/4 - π/б + arctan(б-1/(1+б))
Now we can find the length of the interval by subtracting the endpoints:
π/4 - π/б = (π/4б)(4-б)
Therefore, the average value of csc^2 x over the given interval is:
[π/4 - π/б + arctan(б-1/(1+б))] / [(π/4б)(4-б)]
To determine which of the given choices is the closest to this value, we need to plug in values of б and compare the results. We can start by testing the middle value, б = 2:
[π/4 - π/2 + arctan(3/5)] / [(π/8)(2)] ≈ 2.796
This is closest to choice (C), so our final answer is:
Average of csc^2 x over the interval from x = π/б tо x = π/4 is approximately 2.796.
We can solve this problem by using the definite integral formula for the average value of a function over an interval. The average value of a function f(x) over the interval [a, b] is given by:
avg = (1/(b-a)) * ∫[a, b] f(x) dx
In this problem, we need to find the average value of the function f(x) = csc² x over the interval [π/6, π/4]. Therefore, we have:
avg = (1/(π/4 - π/6)) * ∫[π/6, π/4] csc² x dx
Simplifying the expression inside the integral using the identity csc² x = 1 + cot² x, we get:
avg = (1/(π/12)) * ∫[π/6, π/4] (1 + cot² x) dx
Integrating the first term using the identity ∫csc² x dx = -cot x + C and the second term using the substitution u = cot x, du = -csc² x dx, we get:
avg = (1/(π/12)) * [-cot x - ln|sin x|] [π/4, π/6] + (1/(π/12)) * ∫[cot(π/4), cot(π/6)] (1 + u²) du
Simplifying the first term, we get:
avg = (1/(π/12)) * [sqrt(3)/3 + ln(2)] + (1/(π/12)) * ∫[sqrt(3), sqrt(3)/3] (1 + u²) du
Integrating the second term and simplifying, we get:
avg = (1/(π/12)) * [sqrt(3)/3 + ln(2) + (2sqrt(3))/3 - (2sqrt(3))/9]
avg = (1/(π/12)) * [(4sqrt(3))/9 + ln(2)]
avg = 2.796 (approx.)
Therefore, the correct answer is (C) 2.796.
avg = (1/(b-a)) * ∫[a, b] f(x) dx
In this problem, we need to find the average value of the function f(x) = csc² x over the interval [π/6, π/4]. Therefore, we have:
avg = (1/(π/4 - π/6)) * ∫[π/6, π/4] csc² x dx
Simplifying the expression inside the integral using the identity csc² x = 1 + cot² x, we get:
avg = (1/(π/12)) * ∫[π/6, π/4] (1 + cot² x) dx
Integrating the first term using the identity ∫csc² x dx = -cot x + C and the second term using the substitution u = cot x, du = -csc² x dx, we get:
avg = (1/(π/12)) * [-cot x - ln|sin x|] [π/4, π/6] + (1/(π/12)) * ∫[cot(π/4), cot(π/6)] (1 + u²) du
Simplifying the first term, we get:
avg = (1/(π/12)) * [sqrt(3)/3 + ln(2)] + (1/(π/12)) * ∫[sqrt(3), sqrt(3)/3] (1 + u²) du
Integrating the second term and simplifying, we get:
avg = (1/(π/12)) * [sqrt(3)/3 + ln(2) + (2sqrt(3))/3 - (2sqrt(3))/9]
avg = (1/(π/12)) * [(4sqrt(3))/9 + ln(2)]
avg = 2.796 (approx.)
Therefore, the correct answer is (C) 2.796.
NESTEA APPLE
Forum Guru
letter A
ps: Ang sagot ay (A) 8π dahil ang pagkakalkula gamit ang washer method at ang mga limit ng integrasyon ay nagreresulta sa isang volume na (200/3)π, na halos 209.44 cubic units. Ang tanging opsyon na pinakamalapit sa sagot na ito ay (A) 8π, na katumbas ng 25.13 cubic units. Kaya't ang tamang sagot ay (A) 8π.
ps: Ang sagot ay (A) 8π dahil ang pagkakalkula gamit ang washer method at ang mga limit ng integrasyon ay nagreresulta sa isang volume na (200/3)π, na halos 209.44 cubic units. Ang tanging opsyon na pinakamalapit sa sagot na ito ay (A) 8π, na katumbas ng 25.13 cubic units. Kaya't ang tamang sagot ay (A) 8π.
Last edited:
NESTEA APPLE
Forum Guru
pa mark as solution paps hehe kung oks lang
To find the volume of the solid generated by revolving the region bounded by the parabola, y = -x^2+6x-8, and the x-axis about the y-axis, we can use the method of cylindrical shells.
First, we need to find the values of x where the parabola intersects the x-axis. To do this, we set y = 0 and solve for x:
0 = -x^2+6x-8
x^2 - 6x + 8 = 0
(x-2)(x-4) = 0
x = 2 or x = 4
So the region we need to revolve about the y-axis is bounded by the parabola and the x-axis between x = 2 and x = 4.
Next, we need to find an expression for the radius of the cylindrical shells. Since we are revolving about the y-axis, the radius of each shell will be the distance from the y-axis to the curve at a given value of y. We can express this distance as x = f
, where f is the inverse function of y = -x^2+6x-8.
To find the inverse function, we switch the roles of x and y in the equation:
x = -y^2 + 6y - 8
Then we solve for y in terms of x:
y^2 - 6y + (x+8) = 0
Using the quadratic formula, we get:
y = (6 ± sqrt(36-4(x+8)))/2
y = 3 ± sqrt(16-x)
Since we are revolving about the y-axis, we want the positive square root:
y = 3 + sqrt(16-x)
So the radius of each cylindrical shell is r = 3 + sqrt(16-y).
Finally, we can set up the integral to find the volume of the solid:
V = ∫(2 to 4) 2πr h dy
where h is the height of each cylindrical shell, which is simply the length of the interval we are revolving (in this case, h = 4-2 = 2).
Plugging in the expression for r, we get:
V = ∫(2 to 4) 2π(3 + sqrt(16-y)) 2 dy
V = ∫(2 to 4) 4π(3 + sqrt(16-y)) dy
V = 4π ∫(2 to 4) (3 + sqrt(16-y)) dy
V = 4π [3y + 2/3 (16-y)^(3/2)](2 to 4)
V = 4π [(12 + 8√2/3) - (6 + 8/3)]
V = 4π (6 + 8√2/3)
V = 24π/3 + 32π/9
V = 8π + 32π/9
V = 80π/9
Therefore, the volume of the solid is 80π/9, which is closest to option D, 11π.
First, we need to find the values of x where the parabola intersects the x-axis. To do this, we set y = 0 and solve for x:
0 = -x^2+6x-8
x^2 - 6x + 8 = 0
(x-2)(x-4) = 0
x = 2 or x = 4
So the region we need to revolve about the y-axis is bounded by the parabola and the x-axis between x = 2 and x = 4.
Next, we need to find an expression for the radius of the cylindrical shells. Since we are revolving about the y-axis, the radius of each shell will be the distance from the y-axis to the curve at a given value of y. We can express this distance as x = f
, where f is the inverse function of y = -x^2+6x-8.To find the inverse function, we switch the roles of x and y in the equation:
x = -y^2 + 6y - 8
Then we solve for y in terms of x:
y^2 - 6y + (x+8) = 0
Using the quadratic formula, we get:
y = (6 ± sqrt(36-4(x+8)))/2
y = 3 ± sqrt(16-x)
Since we are revolving about the y-axis, we want the positive square root:
y = 3 + sqrt(16-x)
So the radius of each cylindrical shell is r = 3 + sqrt(16-y).
Finally, we can set up the integral to find the volume of the solid:
V = ∫(2 to 4) 2πr h dy
where h is the height of each cylindrical shell, which is simply the length of the interval we are revolving (in this case, h = 4-2 = 2).
Plugging in the expression for r, we get:
V = ∫(2 to 4) 2π(3 + sqrt(16-y)) 2 dy
V = ∫(2 to 4) 4π(3 + sqrt(16-y)) dy
V = 4π ∫(2 to 4) (3 + sqrt(16-y)) dy
V = 4π [3y + 2/3 (16-y)^(3/2)](2 to 4)
V = 4π [(12 + 8√2/3) - (6 + 8/3)]
V = 4π (6 + 8√2/3)
V = 24π/3 + 32π/9
V = 8π + 32π/9
V = 80π/9
Therefore, the volume of the solid is 80π/9, which is closest to option D, 11π.