# Calc - finding area bounded by curve

25,839 results

**Calc - finding area bounded by curve**

Find the area bounded by x=cubed root of y, y=2, y= -1, and y-axis.

**Calc**

Find the area in the first quadrant bounded by the curve y=(9-x)^1/2 and the x- and y-axis.

**Calc**

Find the area in the first quadrant bounded by the curve y=(9-x)^1/2 and the x- and y-axis.

**AP Calc**

The curve y=x^3 intersects the line y=7x-6 at three points, (-3,-27), (1,1), and (2,8). Find the total area bounded by y=x^3 and y=7x-6.

**AP Calc**

The curve y=x^3 intersects the line y=7x-6 at three points, (-3,-27), (1,1), and (2,8). Find the total area bounded by y=x^3 and y=7x-6.

**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

**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.

**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 ...

**Calculus AB: Area Between Curves**

Hello! I'm having trouble understanding how I'm supposed to work out this problem. Any help would be appreciated! Find the area of the region bounded by the curve y = f(x) = x3 – 4x + 1 and the tangent line to the curve y = f(x) at (–1,4).

**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...

**AP Calc**

Find the point on the curve x=4y-y^2 where the tangent to the curve is a vertical line. My work: Finding the derivative. 1=4(dy/dx)-2y(dy/dx) 1=dy/dx(4-2y) dy/dx=1/4-2y Therefore, y cannot equal +2 or -2 Right?

**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...

**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...

**Maths**

A curve has equation y = 10 + 8x + x^2 - x^3. X >= 0 a)Find the coordinates of the turning point. and show whether it is maximum or minimum. b) Hence, Find the area of the region bounded by the curve, the line y = 11x and the y axis? I manage to complete a) which I got : dy...

**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?

**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...

**Calculus**

The curve y = 11x - 24 - x^2 cuts the x-axis at points A and B, and PN is the greatest positive value of the y coordinate. Show that 2 PN • AB equals three times the area bounded by that portion of the curve which lies in the first quadrant. Kinda confused on how to do this

**Physics**

The curve y = 11x - 24 - x^2 cuts the x-axis at points A and B, and PN is the greatest positive value of the y coordinate. Show that 2 PN • AB equals three times the area bounded by that portion of the curve which lies in the first quadrant. Kinda confused on how to do this

**Calc**

What is the area bounded by the graphs of y = 4x - x² and y = x? Thank you!!

**Calculus, Physics**

The curve y = 11x - 24 - x^2 cuts the x-axis at points A and B, and PN is the greatest positive value of the y coordinate. Show that 2 PN • AB equals three times the area bounded by that portion of the curve which lies in the first quadrant. Kinda confused on how to do this

**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 ...

**Calculus**

(a) Compute the area of the bounded region enclosed by the curve y = e^x, the line y = 12, and the y-axis. (b) How does this area compare with the value of the integral ∫1-12(ln x dx)? Explain your answer. (A picture may be helpful.)

**Calculus**

(a) Compute the area of the bounded region enclosed by the curve y = e^x, the line y = 12, and the y-axis. (b) How does this area compare with the value of the integral ∫ from 1 to 12 of (ln x dx)? Explain your answer. (A picture may be helpful.)

**calculus**

Find the area bounded by the x-axis ,the curve y= 1/x+2, x=0,y=0, and x=5

**maths**

What is the total area bounded by the curve y^2(1-x) = x^2(1+x) and the line x = 1

**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.

**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

**Calculus Homework!**

Find the area bounded by the curve y=x(2-x) and the line x=2y.

**math**

Find the area bounded by the line y = 1 and the curve y = x^2 - 3. Answer 8/3 16/3 32/3 32

**calculus**

Find the area bounded by the curve y=1/2+2, the x-axis, and the lines x=1 and x=4. a.7 1/2 b.10 2/3 c.16 d.28 1/2

**calc**

find the area under the region bounded by the curves y=x^2-3 and y=2x.

**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 ...

**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 ...

**Math calculus-Trig**

Find the area bounded by the curve and the lines y = -x^2 - 4x; y = 1; x = -3; x = 1

**Calc 1**

The region in the first quadrant bounded by the x-axis, the line x = ln(π), and the curve y = sin(e^x) is rotated about the x-axis. What is the volume of the generated solid?

**brief calc**

Calculate the total area of the region described. Do not count area beneath the x-axis as negative.Bounded by the line y = 6x, the x-axis, and the lines x = 4 and x = 5

**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.

**Calculus/ trig**

Find the area bounded by the curve and the lines y=sinx, y= 1/2, x=5pi/6 and x=pi/6

**MATHS**

Find the area of the region bounded by the curve of sin x between x = 0 and x = 2π.

**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 ...

**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 ...

**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 ...

**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.

**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.

**Calc**

Find the value of m such that y=mx divides the area under the curve y=(1-x)x in [0,1] into two regions of equal area D: D: D:

**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 ...

**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.

**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?

**math**

using the concept of the walli's formula find the area bounded by the curve y=sinx and y=cosx from pi/4 to 5pi/4??

**Calc**

R is the region below the curve y=x and above the z-axis from x=0 to x=b, where b is a positive constant. S is the region below the curve y=cos(x) and above the x-axis from x=0 to x=b. For what value of b is the area of R equal to the area of S? I know the answer is 1.404, but...

**calc**

Set up the simplified integral and compute the volume created when the area bounded by y=x^2-1 and y=3 is rotated around the y-axis.

**Calculus**

Finding area under a curve From [1,3] 2x^2 -4x +1

**calc**

The region in the first quadrant bounded by the x-axis, the line x = π, and the curve y = sin(sin(x)) is rotated about the x-axis. What is the volume of the generated solid?

**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, ...

**ap calc**

find the volume generated by revolving the area bounded by y=-x^3, y=0, and x=-2 about the line x=1. I know how to the solve the problem but am not sure how to get the interval. Please help.

**CALC**

Find the area of the region bounded by: y=2x−tan0.18x, x=1, x=5, y=0 Round your answer to 3 decimal places. I get 9 but its wrong! please help

**calc**

Find the maximum possible area of a rectangle in quadrant 1 under the curve y=(x-4)^2 (with one corner at the origin and one corner on the curve y=(x-4)^2)

**Calc **

The figure (which is not pictured, sorry) shows the slope field for y =.2(9-y^2). Make a sketch of the solution using the initial value y(-4)=-2. Find the function f(x) that this solution curve is approaching as x-->infinity. I need help with finding the function f(x) that ...

**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 ...

**Math**

An area is bounded by the x-axis and the parabola y = 16 - x^2. Use four rectangles of equal width and the midpoint approximation method to estimate the bounded area. Could you please show me how to work out this problem? Thanks!

**Calculus**

Find the volume of the solid generated by rotating about the y axis the area in the first quadrant bounded by the following curve and lines. y=x^2, x=0, y=2.

**calculus**

I'm having trouble on this question: Find the area of the region in the first quadrant that is bounded above by the curve y= sq rt x and below by the x-axis and the line y=x -2.

**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 ...

**math**

i need help finding area under curve of:2y=sqrt(3x),y=4, and 2y+1x=4

**Calculus**

i need help finding area under curve of:2y=sqrt(3x),y=4, and 2y+1x=4

**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?

**math**

Sketch the given region R and then find its area. R is the region bounded by the curve y=1/x^2 and the lines y=x and y=x/8

**Calculus**

Sketch the given region R and then find its area. R is the region bounded by the curve y=1/x^2 and the lines y=x and y=x/8.

**calculus**

A region is bounded in the second quadrant by the curve y = ln(1–x), the line y=3, and the y-axis. Find the area of the region.

**Calculus**

The area bounded by the curve y^2 = 12x and the line x=3 is revolved about the line x=3. What is the volume generated?

**AP Calculus AB**

Which integral gives the area of the region in the first quadrant bounded by the axes, y = e^x, x = e^y, and the line x = 4? The answer is an integral. I know y=e^x has no area bounded, but I don't know how to incorporate it all.

**Calculus**

what is Trapezoidal rule? How does it work in finding the area under the curve(s)?

**Stats**

Need help finding the area under the normal curve between z = -1.0 and z = -2.0?

**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?

**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?

**calc**

The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (3, 1) is a point on the curve?

**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

**Calc 1**

Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area. y = x^−3, 1 ≤ x ≤ 5

**Calc**

Find the area of the region bounded by the curves y2 = x, y – 4 = x, y = –2, and y = 1. So far I have found that the area of the trapezoid which is 13.5. But for the other two areas I cannot find them. They could be: 27/2 22/3 33/2 34/3 14 I believe that it is 14 as the ...

**Calc 1**

Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area. y = 6 sin x, 0 ≤ x ≤ π

**Calc 1**

Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area. y = 5th root(x), 0 ≤ x ≤ 32

**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...

**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.

**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.

**calc**

Each of the regions A, B, and C bounded by f(x) and the x-axis has area 5. Find the value of ∫2 [f(x)+3x^2+2]dx. −4 I know to solve I can find the antiderivative of the equation but im not sure how to do this because of the f(x) in the parentheses.

**calc**

Find the area of the region bounded by the curves y equals the inverse sine of x divided by 4, y = 0, and x = 4 obtained by integrating with respect to y. Your work must include the definite integral and the antiderivative.

**calc**

Find the number a such that the line x = a divides the region bounded by the curves x = y^2 − 1 and the y-axis into 2 regions with equal area. Give your answer correct to 3 decimal places.

**Calculus**

Volume created when the area bounded by the curve y = 1/x, the x-axis, and the lines x = 1 and x = 4 is rotated about: a) the x-axis: 2.356 units^3 Is this correct? b) the line y = 5 I'm not sure how to do this one.

**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

**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.

**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

**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 ...

**AP Calc**

find the area in the first quadrant bounded above by y=sinx, below by the x-axis, and to the right by x=pi/2 is divided into two equal parts by the line x=a. Find a.

**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.") ...

**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 ...

**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...

**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 ...

**Calc**

Find the number b such that the line y = b divides the region bounded by the curves y = 4x2 and y = 16 into two regions with equal area. (Round your answer to two decimal places.)

**ap calc**

Find the area of the shaded region, bounded by the parabola 16y=5x^2+16 and the lines y=0, y=6, and x=5. I broke the figure up into two parts and got 3102/5. It seems like a large answer am I correct? Your help is greatly appreciated.

**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.