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. 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 :)
3 answers
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. 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. 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.
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.
6 answers