# Calculus (Area Between Curves)

59,858 results

**Calculus-Area between curves**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y=4*sqrt(x) , y=5 and 2y+4x=8 please help! i've been trying this problem the last couple days, even asked a TA for help, but i can't arrive at...

**calculus**

consider the area enclosed between the curves f (x) = x2 and g (x) = 4x what is the volume obtained by revolving the area between these two curves around the line y = 20 ? please help, i don't know what to do

**calculus**

consider the area enclosed between the curves f(x)=x^2 and g(x)=4x what is the volume obtained by revolving the area between these two curves around the line y=20 ? please help, i don't know what to do

**Calculus (Area Between Curves)**

Find the area of the region bounded by the curves y^2=x, y-4=x, y=-2 and y=1 (Hint: You'll definitely have to sketch this one on paper first.) You get: a.) 27/2 b.) 22/3 c.) 33/2 d.) 34/3 e.) 14

**CALCULUS**

Sketch the region enclosed by the given curves. y = 4/X y = 16x, y = 1X/16 x > 0 and the area between the curves

**Calculus Area between curves**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 3y+x=3 , y^2-x=1

**Calculus (Area Between Curves)**

Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin(x)cos(x)^2, y=2xcos(x^2) and y=4-4x. You get: a.)1.8467 b.) 0.16165 c.) 0.36974 d.) 1.7281 e.) 0.37859

**math**

consider the area enclosed between the curves f (x) = x2 and g (x) = 4x what is the volume obtained by revolving the area between these two curves around the line y = 20?

**Calculus-Steve**

Find the area of the region between the graphs of f(x)=3x+8 and g(x)=x^2 + 2x+2 over [0,2]. I got 34/3. Calculus - Steve ∫[0,2] (x^2+2x+2) dx = 1/3 x^3 + x^2 + 2x [0,2] = 8/3 + 4 + 4 = 32/3 Why are you taking the antiderivative of x^2 +2x+2 when we are trying to find the...

**Calculus (Area Between Curves)**

Find the area of the region IN THE FIRST QUADRANT (upper right quadrant) bounded by the curves y=sin(x)cos(x)^2, y=2xcos(x^2) and y=4-4x. You get: a.)1.8467 b.) 0.16165 c.) 0.36974 d.) 1.7281 e.) 0.37859 Based on my calculations, I would say that the answer is e.) 0.37859. I ...

**calculus area between three curves**

find the area of the region for 2y=4sqrt(x) , y=3, and 2y+3x=7

**calculus**

area between the curves y^2=4x , 7y=2x+20 , 2x+3y=0

**Calculus (Area Between Curves)**

Find the area of the region bounded by the curves y=x^(-1/2), y=x^(-2), y=1 and y=3. You get: a.) 1/2(sqrt(3)) + 4/3 b.) 2(sqrt(3)) - 8/3 c.) 1/2(sqrt(3) - 32/3 d.) 2(sqrt(3)) - 32/3 e.) 8/3 - 2(sqrt(3))

**calculus**

Find the area between the curves y=x^2 & x=2?

**Calculus**

Find the area between the two curves f(x)= ln(x+1)+2 g(x)= x^2-8x+12

**calculus**

Calculate the area between curves y=e^x and y=-e^x on the interval [0,1]

**integral calculus**

Find the area between the curves y=5x and 2y=5x^2

**calculus **

Find the area of the region between the curves y = x^2 and y = 2/(x^2+1).

**Calculus**

Calculate the area of the bounded region between the curves y^2=x and 3y = -3y + 9 ?

**Calculus Area between curves**

Consider the area between the graphs x+6y=8 and x+8=y2. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals where a= , b=, c= and f(x)= g(x)= I found a, but not b or c. I can't seem to figure out f(x) and g(...

**Calculus**

find the area between the two curves y=1/2x and y=x square root of 1-x^2

**Calculus**

Find the area of the region in the first quadrant between the curves y=x^8, and y=2x^2-x^4

**calculus 1**

find the total area between the curves ( y= 4-x ^2 ) and the x-axis over the intervals [0,4]

**Calculus**

Find the area between the two curves y1= x^2 - 4x + 5 and y2 = 2x - 3, finding the points of intersection algebraically.

**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**

1. Find the area of the region between the curves y=sin(x pi/2) and y=x. 2. Find the area of the region between the curves y=sin(x), y=sin(2x), x=0, and x=pi/2.

**Calculus**

1. Find the area of the region between the curves y=sin(x pi/2) and y=x. 2. Find the area of the region between the curves y=sin(x), y=sin(2x), x=0, and x=pi/2.

**Calculus**

1. Find the area of the region between the curves y=sin(x pi/2) and y=x. 2. Find the area of the region between the curves y=sin(x), y=sin(2x), x=0, and x=pi/2.

**Math**

The curves y=sinx and y=cosx intersects twice on the interval (0,2pi). Find the area of the region bounded by the two curves between the points of intersection.

**Math**

The curves y=sinx and y=cosx intersects twice on the interval (0,2pi). Find the area of the region bounded by the two curves between the points of intersection.

**CALCULUS ONE!**

The graph of sinx and cosx intersect once between 0 and pi/2. What is the angle between the two curves at the point where they intersect? (You need to think about how the angle between two curves should be defined).

**Calculus**

Find the area between the following curves y1= x2−4x+3 y2=−x2+2x+3

**Calculus - please help!**

Suppose that 0 < c < ¥ð/2. For what value of c is the area of the region enclosed by the curves y = cos x, y = cos(x - c), and x = 0 equal to the area of the region enclosed by the curves y = cos(x - c), x = ¥ð, and y = 0? i have no idea how to solve this question.

**Calculus**

Find the area between the curves y=(x^3)-10(x^2)+24x and y=(-x^3)+10(x^2)-24x

**Calculus Check my answer**

Find the area between the curves y= x^2/2 +2 and y = –x – 3 on the interval –4 ≤ x ≤ 4. I got 40/3 is that correct?

**Calculus**

Consider the region in the plane consisting of points (x, y) satisfying x > 0, y > 0, and lying between the curves y=x^2 +1and y=2x^2 −2. (b) Calculate the area of this region.

**calculus**

Consider the curves y = x^2and y = mx, where m is some positive constant. No matter what positive constant m is, the two curves enclose a region in the first quadrant.Without using a calculator, find the positive constant m such that the area of the region bounded by the ...

**Calculus Area between curves**

Evaluate the definite integral: sqrt(8-2x) lower limit=-7 upper limit=0 I got -(1/3)(8-2x)^(3/2) and it was wrong. Please Help! Thanks in advance!

**calculus 2**

Use a graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately) the area of the region bounded by the curves. (Round your answer to two decimal places.) y = 8x^2− 3x, y = x^3−8x+ 2

**maths**

determine the co-ordinate of the poin of intersection of the curves y=x*x and y*y=8x. sketch the two curves and find the area enclosed by the two curves.

**Calculus kinda with the area of two curves**

II don't even know where to start with this can anyone help?!? Find c>0 such that the area of the region enclosed by the parabolas y=x^2-c^2 and y=c^2-x^2 is 270.

**Calculus - Center of Mass**

Find the exact coordinates of the centroid given the curves: y = 1/x, y = 0, x = 1, x = 2. X = 1/Area*Integral from a to b: x*f(x)dx Y = 1/Area*Integral from a to b: [(1/2)*(f(x))^2]dx How do I find the area for this? Once I know that, is this the correct set up? X = 1/Area*...

**calculus**

what is the area of the region bounded by the curves y=x^2 , y=8-x^2 , 4x-y+12=0

**calculus**

What is the area under each of the given curves? y=e^x/2 ; x=0 x=1

**Calculus**

Find the total area bounded by the curves y = x^3 and y = x^5 .

**calculus**

Find the area of the region bounded by the curves y=x^2 & y=2x???

**Calculus**

Find the area of the region bounded by the curves y=x^2 - 2x and y= x + 4

**Calculus**

Find the area bounded by the indicated curves. y=x^2, y=0, x=2

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = −x and y = x/2 for x in [0, 6]

**calculus**

Find the area bounded by {y=x2−4 y=4−x2 • sketch the region described • determine any intersection point(s) for the curves (show work!!) • write out the integral(s) that will calculate the area • determine the area (may use a calculator)

**calculus**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. y=3+√X, y=3+1/5x What is the area?

**integral calculus**

FIND THE AREA OF THE REGION BOUNDED BY THE CURVES Y= X^2 + 4X + 3 AND Y= x-1.

**Calculus**

Find the area of the region enclosed by the given curves: 4x+y^2=9, x=2y

**calculus**

Find the area of region bounded by the curves y=sin(pi/2*x)and y=x^2-2x.

**calculus**

Please i need help with this. Find the area bounded by the curves y0…5 - 4x = 0, 2x - y - 4 = 0

**Calculus**

find the area of the region bounded by the curves y=x^2-1 and y =cos(x)

**calculus**

find the area of the region bounded by the curves f(x)=x-x^3 ; g(x)=x^2-x ; over [0,1]

**calculus**

find the area of the regions enclosed by the lines and curves. x+y^2=0 and x+3y^2=2 i know the answer is 8/3, but i don't know how to get it.

**Calculus**

Find the area of the region enclosed by the given curves: y=e^6x, y=2sin(x), x=0, x=pi/2

**Calculus**

Find the area of the enclosed by the y-axis and the curves y=x^2 and y=(x^2+x+1)*e^(-x). ...I'm supposed to use tabular method to find the area. But I'm not sure where to start with this question. I drew the graph for y=x^2, but that's about it. Help! thanks

**Calculas**

area between the curves y=4-x^2 and y=x-8

**Calculus**

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region. 2y = 4(x^(1/2)), y=4, and 2y +2x = 6 I keep getting an area around 21.3 but it is incorrect. Am I close? Thank you!

**calculus**

find the area of one of the regions bounded by the curves y=sin x and y=cos x

**calculus**

How do you find the ariea between these curves? y=4x^2 y=7x^2 4x+y=3 x>=0?

**Calculus - MathMate Please help**

I asked this question on last thursday and u told me that the bottom limit for the 1st and the right limit for the 2nd area is missing? The question that i posted had no other information and i don't know why you said that the limits are missing. There is no other limits that ...

**calculus**

two curves are orthogonal at a point of intersection of their tangents at that point cross at right angles. Show that the curves 2x^2+3y^2=5 and y^2=x^3 are orthogonal at (1,1) and (1,-1). Use parametric mode to draw the curves and to show the tangent lines

**calculus 2**

Sketch the region enclosed by the given curves. Find its area. y=7cos pi x, y=12x^2-3

**calculus**

Find the angle of intersection between the curves f(x) = 2x + 3 and g(x ) = - 3x + 1

**Calculus**

Find the area of the region bounded by the curves y=12-x^2 and y=x^2-6. Hint:The answer should be a whole number.

**Math**

Determine the area of the region between the curves: f(x)= 3x^2 + 16x + 28 and g(x) = -2x + 1

**Math**

Find the area between the two curves y=x^4-10x^2+36 and y=3x^2

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. (Round your answer to three decimal places.) Between y = e^x and y = x for x in [1, 2]

**calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = x2 − 4x + 1 and y = −x2 + 4x − 5 for x in [0, 3]

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = x^2 − 4x + 1 and y = −x^2 + 4x − 5 for x in [0, 3]

**calculus**

find the area enclosed by the given curves. y=squareroot of x y=x/2 x=9 * i just graph it but do not know how to set it up the equation or anything.

**Calculus**

find area of the region bounded by the curves y=x^2-1 and y=cos(x). give your answer correct to 2 decimal places.

**Brief Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = −x and y = −x^3 for x in [−1, 1]

**math**

Determine the area of the region between the curves: f(x) = -(3)x^3 + (6.25)x^2 + (27.1875)x - (9.328125) and g(x) = (1)x^2 + (3)x -(3)

**Math**

Find the area of the region between the curves y=lnx and y=ln2x from x=1 and x=5.

**calculus**

Determine the exact value for the constant k such that the area of the region bounded by the curves y=x and y=kx^2 is equal to 2/3 sq units. Any help is appreciated.

**Calculus**

3) Find the area bounded by the curves f(x)= x^3 + x^2 and g(x)= 2x^2 + 2x. I get 4653pi/105 after help from Mr. Reiny, but this is wrong. Have I calculated incorrectly? Thank you

**Calculus AP**

areas between the curves x=y^2-2, x=y, y=3. int[a,b](f(y)-g(y)) dy i know how to graph. but im have problem with integral.

**Math**

Determine the area of the region between the curves: f(x) = -3x^3 + 6.25x^2 + 27.1875x - 9.328125 and g(x) = 1x^2 + 3x -3

**Maths**

Find the physical area between the curves y=sin(2x) and y=cos(x) from x=pi/2 to x=3pi/2

**Calculus**

Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. Between y = x^2 − 4x + 1 and y = −x^2 + 4x − 5 for x in [0, 3] Please don't...

**Calculus**

Find the number b such that the line y = b divides the region bounded by the curves y = x2 and y = 4 into two regions with equal area.

**Math - Calculus III**

Find the double integral of f (x, y) = (x^7)y over the region between the curves y = x^2 and y = x(3 - x).

**university mathematics**

find the area enclosed between the following curves:- Y= lnx and y= -2x+3 and the ordinate x=3

**University Maths**

Find the area enclosed between the ffg curves :- y=lnx and y= -2x+3 and the ordinate x=3

**calculus wxmaxima**

does anyone know how to approximate the are and circumfrence of the region bounded by the given curves y=cos(x^2 +(100493/100000)), y=1+x-X^2 i already did the area but i need help with the circumference

**Calculus 3**

Given the function f(x,y)=y/(x^2+y^2), give and identify the level curves for k=1, k=1/2, and k=1/4, and draw a contour map showing these level curves. Any help would be immensely appreciated. Thanks

**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...

**math, calculus**

1. Consider the region bounded by the curves y=|x^2+x-12|, x=-5, and x=5 and the x-axis. A. Set up a sum of integrals, not containing an absolute value symbol, that can be used to find the area of this region. B. Find the area of the region by using your answer from part A. ...

**CALCULUS PLEASE HELP**

The birth rate of a population is b(t) = 2300e0.024t people per year and the death rate is d(t)= 1450e0.018t people per year, find the area between these curves for 0 ≤ t ≤ 10

**calculus 2**

The birth rate of a population is b(t) = 2000e^0.022t people per year and the death rate is d(t)= 1460e^0.017t people per year, find the area between these curves for 0 ≤ t ≤ 10. (Round your answer to the nearest integer.)

**Calculus-check my answers please**

dy/dx= (y^2 -1)/x 1. Give the general equation of the curves that satisfy this equation. 2. Show that the straight lines y=1 and y=-1 are also solutions 3. Do any of the curves you found in 1) intersect y=1? My Ans: 1. The general solution i found out to be x^2 + C = (y-1)/(y+...

**Integral calculus**

Use the cylindrical shell method to find the volume of the solid generated by revolving the area bounded by the given curves (x-3)^2 + y^2 = 9, about y-axis.

**math-calculus 2**

Consider the given curves to do the following. 64 y = x^3, y = 0, x = 4 Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about y = 1.

**Calculus**

Find the area of the region which is bounded by the polar curves theta =pi and r=2theta 0<theta<1.5pi inclusive

**Calculus**

Find the area of the region which is bounded by the polar curves theta =pi and r=2theta 0<theta<1.5pi inclusive