Bc 98 the Function F is Continuous on the Closed Interval 2 8

sonofjohn

The function f is continuous on the closed interval [2,8] and has va;ues that are given in the table below. Using sub intervals [2,5]. [5,7], and [7,8], what is the trapezoidal approximation of the anti derivative from 2 to 8 of f(x)dx?
|x| |2|5|7|8|
f(x)| |10|30|40|20|

(a) 110 (b) 130 (c) 160 (d) 190 (e) 210

trapezoid rule= anti derivative from a to b is f(x)dx=T_n= change of x/2*[f(x₁) + 2f(x₂) + ... + 2f(x_n-1) + f(x_n)]

I tried using the formula and plugging the information in exluding the point at [7,40] because it doesn't have the same change in x as the other numbers. Although the equation doesn't specify what the interval count is but rather that I have to use all the intervals. Therefore I don't really know how to use the 3rd interval [7,40] in the problem.

Dick

The usual statement of the trapezoid rule applies to equally spaced intervals. But that doesn't mean you can't apply it if they are not equally spaced. Draw the three trapezoids, find their area and add them.

sonofjohn

Thanks, worked perfectly got 160 units for the area.

sonofjohn

Well this packet I'm doing is quite difficult for me. The semester final is coming up soon and I want to know what I'm doing as best I can. I've decided to scan the first page and see if anyone knows what I'm doing wrong. I've already completed #1 thanks to Dick, but I still can't get 2, 3, or 5.

mutton

2. You assumed that dy/dx = 1/2 (in fact it should be 1) when x = 2 as well, but this is not known. Notice that when x and y change, dy/dx changes too. So you must get rid of dy/dx to get an equation with x and y only.

mutton

3. -cos(3x) is not an antiderivative of sin(3x). Differentiate -cos(3x) to see that you are off by a factor.

5. Same idea as 3, and you know the formula.

sonofjohn

2. You assumed that dy/dx = 1/2 (in fact it should be 1) when x = 2 as well, but this is not known. Notice that when x and y change, dy/dx changes too. So you must get rid of dy/dx to get an equation with x and y only.

how would I go about getting rid of dy/dx?

sonofjohn

3. -cos(3x) is not an antiderivative of sin(3x). Differentiate -cos(3x) to see that you are off by a factor.
5. Same idea as 3, and you know the formula.

All right because of the chain rule, the antiderivative of sin(3x) is -1/3(cos3x). Therefore the final answer comes out to be 2/(3pi). As with number 3, I know the formula for volume but not the formula for mass. The volume formula is pi(r₂)h but how can I fund the mass?

mutton

how would I go about getting rid of dy/dx?

When you want to find dy/dx, you differentiate. What's the "reverse" operation?

5. They already give you the density, so you don't need to find the mass.

sonofjohn

When you want to find dy/dx, you differentiate. What's the "reverse" operation?
5. They already give you the density, so you don't need to find the mass.

If I take the derivative of #2 I get x⁴/8 + x\2, when I plug y=2 into the function. But if I plug 1 into that, I get 5/8. Shouldn't I get 0? And from there plug the new x=2 into the function I just took the antiderivative of?

They give me the density and the height of the rod being 4m. I don't understand what it means when it says x meters from the end of the rod. They gave me the height or length of the rod so are they saying that the diameter of the rod is x? And from there how would I get an average density? Is it like the average value formula I used in question 3?

mutton

2. y isn't always 2, so it doesn't make sense to plug that in before integrating. It may help to rewrite the equation as yy' = x^3 + 1. The fact that y = 2 when x = 1 will let you find the constant of integration.

5. The top of the diagram you drew is one end of the rod. x is not the diameter, but the distance along the length of the rod (vertical distance in your diagram). Since the rod has length 4m, x goes from 0 and 4. Yes to your last question.

sonofjohn

2. y isn't always 2, so it doesn't make sense to plug that in before integrating. It may help to rewrite the equation as yy' = x^3 + 1. The fact that y = 2 when x = 1 will let you find the constant of integration.
5. The top of the diagram you drew is one end of the rod. x is not the diameter, but the distance along the length of the rod (vertical distance in your diagram). Since the rod has length 4m, x goes from 0 and 4. Yes to your last question.

Ok that explained 5 much more clearly. I plugged in the everage formula and got 4 for the average density of the rod, because 1/4* the antiderivative of (2x + x²/2) from 0 to 4 results in 4.

Do I plug the x and y values y=2 when x=1 before or after I take the antiderivative.

mutton

Do I plug the x and y values y=2 when x=1 before or after I take the antiderivative.

If you plug before, what would you be integrating?

sonofjohn

I would be integrating the constant function at y=2 when x=1.

mutton

But that wouldn't make sense. If you already knew that y is constant, the answer would be 2.

sonofjohn

But that wouldn't make sense. If you already knew that y is constant, the answer would be 2.

Yeah, so if I take the antiderivative of the function (x₃ + 1)y I should get y² = x⁴ + x Then from there what do I do?

mutton

First, let's check that what you got is an antiderivative (there are many antiderivatives, you want the one such that y = 2 when x = 1).

[tex]y^2 = x^4 + x[/tex]
[tex]2yy' = 4x^3 + 1[/tex] by differentiation
[tex]y' = \frac{4x^3 + 1}{2y}[/tex] which is not the given equation.

sonofjohn

First, let's check that what you got is an antiderivative (there are many antiderivatives, you want the one such that y = 2 when x = 1).
[tex]y^2 = x^4 + x[/tex]
[tex]2yy' = 4x^3 + 1[/tex] by differentiation
[tex]y' = \frac{4x^3 + 1}{2y}[/tex] which is not the given equation.

I see I took the antiderivative wrong. I don't really know how to take the antiderivative of yy¹ = (x³ - 1)\y, because I don't know how to take the antiderivative of yy¹

mutton

[tex]\int yy' dx = \frac{1}{2} y^2[/tex] since [tex]\frac{d}{dx} y^2 = 2yy'[/tex] by the chain rule.

Maybe you learned some other methods that are helpful here.

sonofjohn

[tex]\int yy' dx = \frac{1}{2} y^2[/tex] since [tex]\frac{d}{dx} y^2 = 2yy'[/tex] by the chain rule.
Maybe you learned some other methods that are helpful here.

alright so 1/2y² = x⁴/4 +x. From there do I plug the x=1 and y=2 in? Or do I plug the x=2 in?

mutton

There's always a constant of integration whenever you evaluate indefinite integrals.

[tex]\frac{1}{2} y^2 = \frac{1}{4} x^4 + x + C[/tex]

Use x = 1, y = 2 to find C.

sonofjohn

There's always a constant of integration whenever you evaluate indefinite integrals.
[tex]\frac{1}{2} y^2 = \frac{1}{4} x^4 + x + C[/tex]

Use x = 1, y = 2 to find C.

Ok so I found C as 3/4. Do I plug x=2 into 1/2y² = 1/4x⁴ + x + 3/4 now?

Bc 98 the Function F is Continuous on the Closed Interval 2 8

Calc 1 trapezoidal approximation

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