Friday

April 18, 2014

April 18, 2014

Number of results: 32,246

**Calc - finding area bounded by curve**

Find the area bounded by x=cubed root of y, y=2, y= -1, and y-axis.
*Thursday, May 26, 2011 at 2:41pm by Amy*

**brief calc**

Calculate the total area of the region described. Do not count area beneath the x-axis as negative. Bounded by the curve y = square root of x the x-axis, and the lines x = 0 and x = 16 This is under Integrals, i don't know what i'm doing wrong, please help
*Wednesday, November 6, 2013 at 9:05pm by kyle*

**Calc help**

Below assumes: y = e^2x means e^(2x) and y = e^5x means e^(5x) The two curves (equations) intersect at x=0, y=1. Then, going in the positive x direction, they diverge. But, the right side of the region is bounded by a line at x=1. So, find the area from x = 0 to x = 1 under ...
*Thursday, February 19, 2009 at 2:16am by Quidditch*

**calculus**

Compute the area of the region in the fi…rst quadrant bounded on the left by the curve y = sqrt(x), on the right by the curve y = 6 - x, and below by the curve y = 1.
*Saturday, January 31, 2009 at 4:01pm by bob*

**calculus help lttle question**

find the area of the regin bounded by the graphs of y=-x^2=2x=3 and y=3. i don't need help solving the problem and but i am a little confused. ok the graph is a parabola and i drew a parobla with y= 3. now when find the area, am I finding the area on the top, above y=3 or on ...
*Monday, January 8, 2007 at 8:31pm by david*

**Calculus**

This is another textbook number that doesn't have the solution and I can't figure it out. Any tips would be greatly appreciated. For each of the plane surfaces, calculate the exact surface area. (Answer in fractions) (a)The surface composed of all surfaces bounded by the curve...
*Thursday, August 9, 2012 at 12:48pm by Paul*

**Calculus**

Find the area of the region bounded by the line y=3x and y= x^3 + 2x^2? and find the area of the region bounded by the curve y=e^2x -3e^x + 2 and the x-axis?
*Tuesday, May 10, 2011 at 7:26pm by Akansha*

**calculus**

Need more info. The curve is one boundary. Is the region to revolve bounded by y=0 and x=4, or by y=0 and x=0, or what? Just revolving a curve will produce a surface, but with no corresponding area, there's no volume.
*Tuesday, December 13, 2011 at 1:53pm by Steve*

**Calculus**

1. Find the area of the region bounded by the curves and lines y=e^x sin e^x, x=0, y=0, and the curve's first positive intersection with the x-axis. 2. The area under the curve of y=1/x from x=a to x=5 is approximately 0.916 where 1<=a<5. Using your calculator, find a. 3...
*Wednesday, February 27, 2013 at 11:17am by Jessy152*

**Calculus**

This is a question from my textbook that does't have a solution and quite frankly I have no idea what to do. Any tips would be greatly appreciated. Given the function f defined by f(x) = 9 - x^2. Find the surface area bounded by the curve y = f(x), the x axis and the lines x...
*Wednesday, August 8, 2012 at 1:18pm by Paul*

**calculus**

Find the area bounded by the x-axis ,the curve y= 1/x+2, x=0,y=0, and x=5
*Tuesday, February 15, 2011 at 5:21pm by jessica*

**maths**

What is the total area bounded by the curve y^2(1-x) = x^2(1+x) and the line x = 1
*Thursday, February 21, 2013 at 8:29am by Anonymous*

**Calc**

What is the area bounded by the graphs of y = 4x - x² and y = x? Thank you!!
*Wednesday, February 2, 2011 at 7:53pm by Erica*

**Math: Need Answer to study for a quizz. Help ASAP **

Estimate the area under the curve f(x)=x^2-4x+5 on [1,3]. Darw the graph and the midpoint rectangles using 8 partitions. Show how to calculate the estimated area by finding the sum of areas of the rectangles. Find the actual area under the curve on [1,3] using a definite ...
*Thursday, November 18, 2010 at 7:35pm by Jesse*

**calculus**

How do I find the area bounded by the curve y = x^1/2 + 2, the x-axis, and the lines x = 1 and x = 4
*Tuesday, January 25, 2011 at 2:49pm by Teri*

**Calc**

Find the area of the region bounded by y=x^2 and y = -(x-4)^2 +4 and the lines y=0 and y=4.
*Thursday, April 2, 2009 at 1:10am by Jose*

**Calculus Homework!**

Find the area bounded by the curve y=x(2-x) and the line x=2y.
*Thursday, May 9, 2013 at 7:32pm by Prue *

**calc**

find the area of the region bounded by y=4x, y=x^3, x=0 and x=2 Check this on paper, but in my mind I see the differential area (x^3-3x)dx integrate that from 0 to 2 y=4x and y=x^3 intersect when x=-2,0 and 2 the vertical boundaries are x=0 and x=2, so the area is the integral...
*Wednesday, May 2, 2007 at 4:15pm by quita*

**Math calculus-Trig**

Find the area bounded by the curve and the lines y = -x^2 - 4x; y = 1; x = -3; x = 1
*Thursday, April 21, 2011 at 6:16pm by Tina*

**MATHS**

Find the area of the region bounded by the curve of sin x between x = 0 and x = 2π.
*Tuesday, November 27, 2012 at 12:53am by CASEY*

**Calculus/ trig**

Find the area bounded by the curve and the lines y=sinx, y= 1/2, x=5pi/6 and x=pi/6
*Thursday, April 21, 2011 at 5:17pm by Rea*

**math**

find the area of a rectangle bounded by the axes and one of its corner is a point in the curve y=1/x.
*Monday, July 16, 2012 at 10:25pm by patrick*

**Math**

Estimate the are under the curve f(x)=x^2-4x+5 on [1,3]. Darw the graph and the midpoint rectangles using 8 partitions. Show how to calculate the estimated area by finding the sum of areas of the rectangles. Find the actual area under the curve on [1,3] using a definite integral.
*Thursday, November 18, 2010 at 7:17pm by Jesse*

**math**

The region R, is bounded by the graphs of x = 5/3 y and the curve C given by x = (1+y^2)^(1/2), and the x-axis. a) Set up and evaluate an integral expression with respect to y that gives the area of R. b) Curve C is part of the curve x^2 - y^2 = 1, Show that x^2 - y^2 = 1 can ...
*Sunday, March 2, 2008 at 2:56pm by leess*

**math**

The region R, is bounded by the graphs of x = 5/3 y and the curve C given by x = (1+y^2)^(1/2), and the x-axis. a) Set up and evaluate an integral expression with respect to y that gives the area of R. b) Curve C is part of the curve x^2 - y^2 = 1, Show that x^2 - y^2 = 1 can ...
*Sunday, March 2, 2008 at 5:20pm by PLEASE HELP!*

**MATH-HELP!**

The region R, is bounded by the graphs of x = 5/3 y and the curve C given by x = (1+y^2)^(1/2), and the x-axis. a) Set up and evaluate an integral expression with respect to y that gives the area of R. b) Curve C is part of the curve x^2 - y^2 = 1, Show that x^2 - y^2 = 1 can ...
*Sunday, March 2, 2008 at 7:48pm by PLEASE HELP!*

**why won't anybody help me**

The region R, is bounded by the graphs of x = 5/3 y and the curve C given by x = (1+y^2)^(1/2), and the x-axis. a) Set up and evaluate an integral expression with respect to y that gives the area of R. b) Curve C is part of the curve x^2 - y^2 = 1, Show that x^2 - y^2 = 1 can ...
*Sunday, March 2, 2008 at 8:29pm by PLEASE HELP!*

**calculus**

find the area of a rectangle bounded by the axes and one of its corner is a point in a curve y=1/x.
*Tuesday, July 17, 2012 at 9:28pm by gracel morales*

**Math**

Find the volume generated when the area bounded by the curve y^2 = 16x, from x = 0 to x = 4 is rotated around the x-axis.
*Tuesday, August 21, 2012 at 1:32pm by Fred*

**math/calc**

Your curve y+x^2 e^(-3x) what is that equal to ? A curve has to follow an equation. If you mean the curve y=x^2 e^(-ex) then the area is yx/2 which is equal to 1/2 x^3 e^(-3x). Right? dArea/dx= 3x^2e^(-3x)-3x^3e^(-3x)=0 solve for x. x=1 check that I did it in my head. Mornings...
*Wednesday, March 31, 2010 at 9:22am by bobpursley*

**Calculus**

Integrals: When we solve for area under a curve, we must consider when the curve is under the axis. We would have to split the integral using the zeros that intersect with the axis. Would this be for all integrals? What if we just want to "find the integral", without finding ...
*Friday, December 18, 2009 at 1:56pm by Jennifer*

**Calc 121**

Okay, how would you go about finding the area of a curve from 1 to 4, when y=2x+(2/(x^2))?? It's not like the problem I asked before because here, you cannot use substitution. I tried using 2x for u and x^2 for du but it won't simplify into a ln problem or anything that I can ...
*Sunday, April 22, 2007 at 4:48pm by Me*

**Calculus**

Finding area under a curve From [1,3] 2x^2 -4x +1
*Thursday, December 8, 2011 at 12:31pm by Tim*

**Maths**

Let a*pi be the area of the region bounded by the curve defined by the parametric equations x=10cos2t, y=10sin2t where 0<=t<=2pi. What is a?
*Tuesday, October 29, 2013 at 2:04pm by Marlen*

**Calculus**

The area bounded by the curve y = 2x^2-x^3 and line y=0 is rotated around the y-axis. The volume of the resulting structure can be expressed as V = a(pi)/b, where a and b are coprime positive integers. What is the value of a + b?
*Friday, March 15, 2013 at 6:22am by Devin*

**Calc 2**

Sure looks to me that you are simply finding the area of the circle x^2 + y^2 = 25 the radius would be 5 so the area is 25π
*Wednesday, September 5, 2012 at 12:27am by Reiny*

**Calc 121**

Okay, how would you go about finding the area of a curve from 1 to 4, when y=2x+(2/(x^2))?? It's not like the problem I asked before because here, you cannot use substitution. I tried using 2x for u and x^2 for du but it won't simplify into a ln problem or anything that I can ...
*Saturday, April 21, 2007 at 10:50pm by Me*

**math:Calculus**

find the area bounded by the curve f(x) =-x^2 +6x -8 and the x axis using both left endpoint and right endpoint summation.
*Wednesday, April 20, 2011 at 8:04am by iqra*

**calculus**

The area bounded by the curve 2y^2=x and the line 4y=x is rotated around the y-axis. The volume of the resulting structure can be expressed as V=a/bπ, where a and b are coprime positive integers. What is the value of a+b?
*Wednesday, July 17, 2013 at 7:28pm by andy*

**calculus**

The area bounded by the curve 2y^2=x and the line 4y=x is rotated around the y-axis. The volume of the resulting structure can be expressed as V=a/bπ, where a and b are coprime positive integers. What is the value of a+b?
*Thursday, July 18, 2013 at 12:14pm by andy*

**Calculus**

what is Trapezoidal rule? How does it work in finding the area under the curve(s)?
*Tuesday, November 30, 2010 at 2:06pm by prakash*

**calculus**

1. Find the area of the region bounded by f(x)=x^2 +6x+9 and g(x)=5(x+3). Show the integral used, the limits of integration and how to evaluate the integral. 2. Find the area of the region bounded by x=y^2+6, x=0 , y=-6, and y=7. Show all work required in #1. 3. Find the ...
*Tuesday, May 15, 2012 at 7:40pm by katarina*

**calculus**

Poorly worded question. Since y = x is a nice straight line going through the origin at a 45° angle, the area "under the curve" is open - ended and would be infinite. If you wanted the area of the region bounded by y = x, y = -3, and x = 3 , you would simply have a right-...
*Tuesday, November 5, 2013 at 2:50am by Reiny*

**Calculus**

The question is find the area of the reagion that is bounded by the curve y=arctan x, x=0, x=1, and the x-axis. So I've drawn the enclosed region. To find the area would I use the Disc/shell method? If so the formula that I came up with looks like this: If area = pi(r)^2 then ...
*Friday, July 12, 2013 at 7:54pm by Isaac*

**calculus**

find the area of the region bounded by the polar curve r=sqrt(6ln(theta)+3 as well as the rays theta=1 and theta=e
*Wednesday, October 13, 2010 at 9:08pm by Bob*

**calculus**

10). Find the area of the region bounded by the graph of f(x)=4lnx/x, y=0, x=5.8 11). Find the area of the region bounded by x=y^2-1y, x=0
*Sunday, April 14, 2013 at 9:20pm by Sally*

**One for Drwls (calc)**

Yes. INT f'dx is f(x). The area under the f'curve is the integral.
*Saturday, September 15, 2007 at 8:19am by bobpursley*

**CALC - area under a curve**

You have an unknown function that is monotone increasing for 1<x<5 and have the following information about the function values. With the clear understanding that there is no way to get an exact integral, how would you try and approximate the area under the curve? X=(1, ...
*Thursday, March 6, 2008 at 6:44pm by anonymous*

**Calc**

Even though they don't intersect, there is the area bounded between y=0 and y=2 Did you notice that the two parabolas are congruent and the second is merely translated 2 units to the right? so the horizontal distance between corresponding points is always 2 that is x2-x1 = (y-...
*Monday, October 10, 2011 at 2:27am by Reiny*

**More Calc**

Find the area between each curve and the x-axis for the given interval. y=6x^2+5 from x=0 to x=5 Thanks.
*Wednesday, May 14, 2008 at 6:12pm by John*

**Calculus ll - Improper Integrals**

Find the area of the curve y = 1/(x^3) from x = 1 to x = t and evaluate it for t = 10, 100, and 1000. Then find the the total area under this curve for x ≥ 1. I'm not sure how to do the last part of question ("find the the total area under this curve for x ≥ 1.") ...
*Saturday, October 2, 2010 at 9:32pm by Alyssa*

**CALC - area under a curve**

area= sum y*deltax start at x=1, make a table deltax yavg .2 (6.675/2=?) .4 (3.505+4.317)/2 and so on, then add the products of yavg*dx
*Thursday, March 6, 2008 at 6:44pm by bobpursley*

**Calc**

Find the area of the largest rectangle that can be inscribed under the curve y = e^(-x^2) in the first and second quadrants.
*Tuesday, December 7, 2010 at 7:44am by Erica*

**math/calc**

You appear to have a typo, I will assume your curve is y = x^2e^(-3x) or y = x^2/e^(3x) let the point of contact of the hypotenuse be (x,y) on the curve. Then the right angle will be at (x,0) and the Area = xy/2 = (1/2)(x)x^2e^(-3x) 2A = (x^3)(e^(-3x)) 2dA/dx = x^3(-3e^(-3x...
*Wednesday, March 31, 2010 at 9:22am by Reiny*

**calculus**

find the volume of the solid of revolution obtained by revolving the region bounded above by the curve y=f(x) = √16-x^2 and below by the curve y=g(x) from x=0 to x=x√2 about the x-axis
*Saturday, March 3, 2012 at 5:01am by Anonymous*

**calculus**

find the volume of the solid of revolution obtained by revolving the region bounded above by the curve y=f(x) = √16-x^2 and below by the curve y=g(x) from x=0 to x=x√2 about the x-axis
*Sunday, March 4, 2012 at 11:52am by help pleaseee*

**Calc**

Find the areas of the regions bounded by the lines and curves by expressing x as a function of y and integrating with respect to y. x = (y-1)² - 1, x = (y-1)² + 1 from y=0 to y=2. I graphed the two functions and the do not intersect? Does it matter? Or do I still find the area...
*Monday, October 10, 2011 at 2:27am by Erica*

**Calculus**

Find the area cut off by x+y=3 from xy=2. I have proceeded as under: y=x/2. Substituting this value we get x+x/2=3 Or x+x/2-3=0 Or x^2-3x+2=0 Or (x-1)(x-2)=0, hence x=1 and x=2 are the points of intersection of the curve xy=2 and the line x+y=3. Area under curve above X axis ...
*Tuesday, April 1, 2014 at 3:07am by MS*

**Ap calc.. Dying!! Please help!**

Given the curve x^2-xy+y^2=9 A) write a general expression for the slope of the curve. B) find the coordinates of the points on the curve where the tangents are vertical C) at the point (0,3) find the rate of change in the slope of the curve with respect to x I don't even know...
*Friday, January 25, 2013 at 9:44pm by Aparna*

**Calc**

First find the intersections between the two curves, they are the limits of integration. They can be found readily as x=±2. Next, we check that the curves do not cross the axis y=-1 between the limits of integration. Otherwise the limits of integration must change. Now ...
*Thursday, December 15, 2011 at 7:15pm by MathMate*

**Calc 2: Area under the curve**

Find the area of the region enclosed between y=2sin(x and y=3cos(x) from x=0 to x=0.4pi Hint: Notice that this region consists of two parts. Notice: I'm getting 1.73762 but apparently that is wrong.
*Wednesday, April 20, 2011 at 3:32am by sam*

**CALC**

If you are finding the area bounded by y=1, y=54.6(approx), x=0, x=4, and f(x,y)=e^x, or any other one-to-one function, and if the limits for ∫e^x dydx are y=1,f(x,y), x=0,4 then changing the order could be: x=e^x to 4, y=1, e^4. However, it depends on the properties of ...
*Wednesday, December 1, 2010 at 11:21pm by MathMate*

**Calc.**

sketch the curve using the parametric equation to plot the points. use an arrow to indicate the direction the curve is traced as t increases. Find the lenghth of the curve for o<t<1. Find an equation for the line tangent to the curve at the point where t=-t. the equation...
*Saturday, April 14, 2007 at 4:03pm by Sammy*

**calc**

Find the area under the curve below from x = 0 to x = 2. Give your answer correct to 3 decimal places. y = 2x - x2
*Thursday, October 28, 2010 at 11:05am by Anonymous*

**calculus**

The region A is bounded by the curve y=x^2-5x+6 and the line y = -x + 3. (a) Sketch the line and the curve on the same set of axes. (b) Find the area of A. (c) The part of A above the x-axis is rotated through 360degree about the x-axis. Find the volume of the solid generated...
*Tuesday, December 11, 2012 at 12:54pm by kchik*

**Calc.**

Find the area of the region bounded by the parabola y=x^2, the tangent line to this parabola at (1,1) and the x-axis. I don't really get what this question is asking. It looks like the area of right triangle to me...try the graph, and shade the area under the tangent. Find out...
*Tuesday, September 26, 2006 at 8:08pm by Hebe*

**calculus**

A region is bounded by the function y=2x^2+3 and the x-axis over the interval(0,2). Sketch the graph of the bounded region. Use the limit process to find the area of the bounded region. Explain the step in this limit process. Please explain all procedured using correct ...
*Sunday, April 10, 2011 at 11:08pm by Anonymous*

**Calculus 2**

Calculus work problem. John bought the following corner lot. Two sides are bounded by 2 roads the back is bounded by a stream that follows the curve y = -x^3 + 8. He wants to build a house in the center of the lot. Where will that be? Please show all work
*Thursday, September 27, 2012 at 5:35pm by Laura*

**math**

i need help finding area under curve of:2y=sqrt(3x),y=4, and 2y+1x=4
*Saturday, March 5, 2011 at 4:48pm by applebottom*

**Calculus**

i need help finding area under curve of:2y=sqrt(3x),y=4, and 2y+1x=4
*Saturday, March 5, 2011 at 5:20pm by applebottom*

**calc**

Let R be the region bounded by the y-axis and the graph of y=xcubed divided by 1+xsquared and y=4-2x,as shown inthe figure above. find the area of R find the volume of the solid generated when R is revolved about the x-axis
*Thursday, April 2, 2009 at 9:52pm by Anonymous*

**math**

Each shape has a different formula for finding its area. This site has formulas for finding the area of some shapes. (Broken Link Removed)
*Monday, January 14, 2008 at 9:47pm by Ms. Sue*

**calc**

Find the area of the surface obtained by rotating the curve of parametric x=3t-(3/3)t^3 y=3t^2 0<=t<=1 what is the surface area
*Thursday, October 30, 2008 at 11:00pm by alexis*

**Calc**

Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the y-axis. y = cube rt. (x) + 2 Thank you so much!!
*Monday, February 7, 2011 at 1:11am by Erica*

**MATH**

(a) Transform the expression (x − a)^2 + y^2 = a^2 into polar coordinates. (b) Sketch the region R bounded by the curve given in part (a). (c) Use a double integral in polar coordinates to ﬁnd the area of the region R.
*Saturday, October 26, 2013 at 3:23am by help please*

**Calculus II**

Here's the problem: Find the area of a plane region bounded by y=x^3 and its tangent line through (1,1). So far I have the graph on my graphing calc, so I have an idea at what I'm looking at. I found the tangent line to y=x^3 to be y=3x-2. Now I am stuck. I am trying to find ...
*Tuesday, May 29, 2007 at 10:59pm by Ace*

**Help Calc.!**

original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point P.
*Wednesday, November 12, 2008 at 6:06pm by Rainie*

**Calc**

Find the coordinates of the point on the curve xy=6 in quadrant I where the normal to the curve passes through the origin.
*Sunday, February 16, 2014 at 12:18pm by Anon*

**calc**

The slope of the tangent to the curve y=f(x) is given by (1/9)(x^2)(y^2). The curve passes through the point (3,1). Find the value of y when x=3*cubedroot(3).
*Sunday, February 16, 2014 at 10:45am by Anon*

**please help; calc**

original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point P.
*Wednesday, November 12, 2008 at 8:01pm by stephanie*

**geometry**

Consider the region in Quadrant 1 totally bounded by the 4 lines: x = 3, x = 9, y = 0, and y = mx (where m is positive). Determine the value of c such that the vertical line x = c bisects the area of that totally bounded region. Needless to say, your first task should be to ...
*Monday, March 18, 2013 at 7:18pm by borat*

**geometry**

Consider the region in Quadrant 1 totally bounded by the 4 lines: x = 3, x = 9, y = 0, and y = mx (where m is positive). Determine the value of c such that the vertical line x = c bisects the area of that totally bounded region. Needless to say, your first task should be to ...
*Monday, March 18, 2013 at 7:25pm by borat*

**math**

Consider the region in Quadrant 1 totally bounded by the 4 lines: x = 3, x = 9, y = 0, and y = mx (where m is positive). Determine the value of c such that the vertical line x = c bisects the area of that totally bounded region. Needless to say, your first task should be to ...
*Tuesday, March 19, 2013 at 6:51pm by borat*

**math/calc**

How do I solve for an area under the curve when y = x^2 and solve for above the interval (1,2) ? Any resources where I could see examples would be helpful. Thanks!
*Wednesday, July 18, 2012 at 11:18am by cathy*

**Economics**

Ok, in your graphs, have both the supply curve and the demand curve touch the Y-axis. Consumer surplus is represented by the area below the demand curve but above equilibrium price. Producer surplus is represented by the area above the supply curve but below equilibrium price...
*Tuesday, March 11, 2008 at 11:04am by economyst*

**calc**

Suppose a curve is traced by the parametric equations x=1sin(t) y=18-3cos^2(t)-4sin(t) At what point (x,y) on this curve is the tangent line horizontal?
*Thursday, October 30, 2008 at 10:58pm by alexis*

**calculus**

Set the two equations equal to each other and solve Go look at your integration chapter - area bounded by two curve Integration (line - parabola)dx Then you are going to deal with adding the part of the parabola under the x-axis
*Tuesday, December 11, 2012 at 12:50pm by Anonymous*

**Math: Calculus**

draw the curve. THe area under the curve from t=0 to t=7 is distance (velocity*time=distance). It should be a triangle, and you can figure the area of that.
*Wednesday, November 10, 2010 at 3:35pm by bobpursley*

**AP Statistics**

A certain density curve looks like an interverted letter V. The first segment goes fro the point (0,0.6) to the point (0.5,1.4). The segment goes from (0.5.1.44) to (1,0.6). (a) Sketch the curve. Verify that the area under the curve is 1, so that it is a valid density. -- Okay...
*Sunday, October 12, 2008 at 10:48pm by Geroge*

**Calc**

Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis. y = (x³/6) + (1/2x), 1≤ x ≤ 2
*Monday, February 7, 2011 at 1:05am by Erica*

**calc ii**

I got it. its that the minimum area is found by saying that the function is a straight line at y=12 and the opposite for 18 and finding those areas under the integrals bounds
*Friday, September 27, 2013 at 8:13pm by Anonymous*

**Calc**

Hi! Please help :-) Find an equation for the circle of the curvature of the curve r(t) = ti + (sint)j at the point (pi/2, 1) (The curve parametrizes the graph of y = sinx in the xy-plane)
*Sunday, December 9, 2007 at 10:44pm by Sadie*

**Calc :(((((**

The curve y = x/(1 + x^2)is called a serpentine. Find an equation of the tangent line to this curve at the point (4, 0.24). Round the slope and y-intercept to two decimal places.)
*Friday, March 7, 2014 at 2:58pm by Tenshi *

**Calculus**

I will assume you want the area between the curve and the x-axis. Under the curve would be infinitely large, it would be open below. area = ∫ x^2 + x) dx from 0 to 3 = [(1/3)x^3 + (1/2)x^2 ] from 0 to 3 = 9 + 9/2 - 0 - 0 = 27/2 square units
*Sunday, February 12, 2012 at 4:35pm by Reiny*

**calculus**

suppose that the area under the curve y=1/x from x=a to x=b is k. the area in terms of k, under the curve y=1/x from x=2a to x=2b? justify
*Tuesday, February 23, 2010 at 5:43pm by Anonymous*

**statistics**

when finding the area to the left of the Z score and Z= 1.43 do i subtract the z score from one or is that only when finding the area to the right of Z
*Sunday, November 6, 2011 at 6:52pm by kate*

**last calc question, i promise!**

given the curve x + xy + 2y^2 = 6... a. find an expression for the slope of the curve. i got (-1-y)/(x + 4y) as my answer. b. write an equation for the line tangent to the curve at the point (2,1). i got y = (-1/3)x + (5/3). but i didn't any answer for c! c. find the ...
*Friday, February 20, 2009 at 12:26am by jane*

**CALC**

Consider half of the rectangle which lies completely in the first quadrant. The lower left corner is the origin (0,0), and the upper right corner lies on the curve y=9-x². The area of the (half) rectangle is therefore: A(x) =x*y =x*(9-x²) =9x-x³ Differentiate ...
*Sunday, May 30, 2010 at 1:27am by MathMate*

**mechanics**

Sketch a stress-strain curve for which the stress is calculated using the actual cross-sectional area rather than the original cross-sectional area. Explain why this curve differtypical stress-strain curve
*Sunday, February 24, 2013 at 12:21pm by dmkp*

**Need help fast on calc**

original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point P. -when plugging y as 2 in original, it ...
*Wednesday, November 12, 2008 at 11:27pm by Craig*

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