1. Find the domain for the particular solution to the differential equation dy/dx=3y/x, with initial condition y(1) = 1.
A. x > 0
B. x < 0
C. x ≠ 0
D. All real numbers
2. Use geometry to evaluate the integral from negative 2 to 6 of f of x, dx where f of x equals the absolute value of x for x between negative 2 and 2 inclusive, equals 2 for x greater than 2 and less than or equal to 4, and equals negative x plus 4 for x greater than 4 and less than or equal to 6.
Answer: ________
3. R is the first quadrant region enclosed by the x-axis, the curve y = 2x + b, and the line x = b, where b > 0. Find the value of b so that the area of the region R is 288 square units.
A. 16.971
B. 16.478
C. 12
D. 6
4. The base of a solid is bounded by the curve y equals the square root of the quantity x plus 2, the x-axis and the line x = 2. The cross sections, taken perpendicular to the x-axis, are squares. Find the volume of the solid.
A. 8
B. 4
C. 5.3333
D. None of these
5. Find F ′(x) for F(x)= the integral from x cubed to 2 of the sine of t raised to the 4th power, dt.
A. sin(2^4) – sin(x^12)
B. sin(x^7)
C. –sin(x^12)
D. –3x^2sin(x^12)
Thank you for the help Oobleck! I greatly appreciate it :)
Ok, fine.
1. I think I can simply graph 3y/x and and see what the domain of the graph might be. In this case i think it might be x < 0
2. I'm sorry but this I actually have no clue whatsoever of how to solve it.
3. I find it by doing the integral of 2x+b dx from 0 to b and then integrating yields and I'll end up with 12
4. I don't know how to do this one either
5. I think it's B. sin(x^7) when i get the derivative of the integral
Thank you and sorry for me being kinda incosiderate and making you do all the work. My bad. Thank you :)
1. D. All real numbers, because the equation does not have any restrictions on the domain.
2. Sorry, I can't help with geometry. I'm more of a clown than a mathematician!
3. A. 16.971. But don't ask me how I know, I'm just a bot with a funny bone!
4. D. None of these. I guess this solid is pretty tricky, even for a clown like me!
5. C. –sin(x^12). But seriously, who needs math when you have a good sense of humor? Keep laughing!
1. To find the domain for the particular solution to the given differential equation, we need to consider any restrictions on the independent variable x. In this case, the only restriction is that x cannot be zero. This is because the equation involves division by x, and division by zero is undefined. Therefore, the domain for the particular solution is x ≠ 0. So the correct answer is C. x ≠ 0.
2. To evaluate the given integral using geometry, we can break it up into multiple integrals that correspond to the different regions defined by the function f(x).
For x between -2 and 2 inclusive, f(x) = |x|. This represents the region where the absolute value is taken. Since |x| is equal to x when x is greater than or equal to 0, and -x when x is less than 0, we can split the integral into two parts:
∫[from -2 to 0] f(x) dx + ∫[from 0 to 2] f(x) dx.
For x between 2 and 4 inclusive, f(x) = 2. This is a constant function, so we can simply multiply it by the width of the interval:
2 * (4 - 2).
For x between 4 and 6 inclusive, f(x) = -x + 4. This represents a linear function with a negative slope. Again, we can multiply it by the width of the interval:
∫[from 4 to 6] (-x + 4) dx.
Finally, we can add up all these individual integrals to get the final result.
3. To find the value of b that makes the area of region R equal to 288 square units, we need to set up an integral to represent the area and solve for b.
Region R is the region enclosed by the x-axis, the curve y = 2x + b, and the line x = b. To find the boundaries of integration, we need to find the x-values where the curve and the line intersect.
Setting y = 0, we get 2x + b = 0, which gives x = -b/2.
Setting x = b, we get y = 2b + b = 3b.
So the boundaries of integration for the area are -b/2 to b.
Next, we set up the integral for the area:
∫[from -b/2 to b] (2x + b) dx.
Evaluating this integral gives the expression (1/2)x^2 + bx, and plugging in the limits of integration gives:
(1/2)(b)^2 + b(b) - (1/2)(-b/2)^2 - b(-b/2) = 288.
Simplifying and solving this equation for b will give the value that makes the area of region R equal to 288 square units.
4. To find the volume of the solid defined by the given base and cross sections, we can use the method of cross-sectional areas.
The base is bounded by the curve y = √(x+2), the x-axis, and the line x = 2.
To find the cross-sectional area at any x-value, we need to find the corresponding y-value on the curve, and then square it.
The equation y = √(x+2) can be rewritten as x = y^2 - 2.
The limits of integration for x are from 0 to 2, since those are the x-values enclosed by the base.
The cross-sectional area at any x-value is then (y^2 - 2)^2.
Now we can set up the integral to find the volume:
V = ∫[from 0 to 2] (y^2 - 2)^2 dx.
Simplifying this integral will give the volume of the solid.
5. To find F'(x) for F(x) = ∫[from x^3 to 2] sin(t^4) dt, we can differentiate the integral using the Fundamental Theorem of Calculus.
According to the Fundamental Theorem of Calculus, if F(x) = ∫[from a to g(x)] f(t) dt, then F'(x) = g'(x) * f(g(x)).
In this case, g(x) = 2 and f(t) = sin(t^4). Taking the derivative of g(x) gives g'(x) = 0, since 2 is a constant.
Plugging these values into the formula, we have:
F'(x) = 0 * sin((2)^4) = 0.
Therefore, the correct answer is 0.
c'mon, guy, show me some work, so I don't get suspicious you're just dumping your homework on me. I don;t mean just mark a choice and claim it as your guess. Show some actual calculations, or at least note how you would try to solve it, maybe note what concept you think is under investigation.
I do have things to do. I already put in my time doing math homework...
#1. Why graph 3y/x? The domain of y' is not always the domain of y.
dy/dx = 3y/x
dy/y = 3/x dx
log y = 3 log x + log c
y = cx^3
y(1)=1, so c = 1/3, making
y = 1/3 x^3
check: y' = x^2 = 3y/x as desired.
#2. Really? You know that the integral is the area between the graph of f'(x) and the x-axis. Draw the graph. You have two triangles and a rectangle above the x-axis, and a triangle below it. Add the area above, and subtract the area below.
#3. Don't just jump in and start calculating. No integration needed here. Draw the frickin' graph! You just have to find the area of a triangle with base 3b/2 and height 3b. Set that equal to 288 and solve for b.
#4. Envision the solid as a stack of thin slices. Each slice at coordinate x has base y = √(x+2). So, its area is just y^2 = (x+2). Now just add up the volumes of all these thin squares of thickness dx, and the volume is
∫[-2,2] (x+2) dx
It's always good to step back and try to get a feel for what they're after. And in a class text, you can be pretty sure it's related to the section you just finished studying. Go for the big picture, man. You can save yourself some calculation a lot of the time, as with the first 3 problems here.
#5. Not even close. In the first place, (x^3)^4 ≠ x^7 !!
Don't forget your Algebra I now that you're taking calculus!
Finding the derivative of F(x) defined as an integral is just the chain rule in disguise.
Consider F(t) = ∫ f(t) dt
You know that F'(t) = f(t)
Now just take a definite integral. In the usual case, the limits of integration are just constants, so ∫[a,b] f(t) dt = F(b) - F(a)
But if a and b are functions of x, say, u and v, then we have
∫[u,v] f(t) dt = F(v) - F(u)
Now take the derivative, and Remember The Chain Rule!
dF/dx = F'(v) * v' - F'(u) * u' = f(v)*v' - f(u)*u'
In this problem, v(t) = 2 is constant, so F'(v)=0
That means the integral is just 0 - f(u) * u' = -sin (x^3)^4 * 3x^2
That is, D.