What is the area enclosed by |x|+|y|=1?
First, obtain the equation of the straight line in each quadrant:1st Quadrant [math](x>0, y>0): x+y=1[/math]2nd Quadrant [math](x<0,y>0): -x+y=1[/math]3rd Quadrant: [math](x<0,y<0): -x-y=1[/math]4th Quadrant: [math](x>0,y<0): x-y=1[/math]Now, plot the combined equations in each quadrant of the rectangular co-ordinate system.The figure appears like this:So, its a rectangle, with length of each side as [math]√2[/math] unit.Consequently, the area enclosed by the function is [math]√2*√2 = 2[/math] square unit.
Let c be the region enclosed by the graphs of y=e^x, y=x, and lines x=0 and x=4?
a) Imagine the area as composed of a bunch of little "dx" rectangles, each having height h = e^x - x. Then A = ∫[0,4] (e^x - x) dx = (e^x - ½x²) |[0,4] = (e^4 - 8 - (1 - 0)) = 54.6 - 9 = 45.6 b) now rotate these rectangles about the x-axis. Each generated "washer" has outer radius R = e^x and inner radius r = x, so V = ∫[a,b] π(R² - r²) dx = π∫[0,4] (e^2x - x²) dx = π(½e^2x - (1/3)x³) |[0,4] V = π(½e^8 - 64/3 - (½ - 0)) ≈ 4614 barring computational error c) Use shells. Each shell will have radius R = x and height h = y2 - y1 = e^x - x, so V = 2π∫[0,4] x(e^x - x) dx
The area of the region enclosed by the graphs of y=x and y=x^2-3x+3 is?
Q: What is the area enclosed by y = x and y = x^2 - 3x + 3 Create a table to see which function is on top f1(1) = 1 f2(1) = 1 f1(2) = 2 f2(2) = (2)^2 - 3(2) + 3 = 4 - 6 + 3 = - 2 + 3 = 1 So f2(x) is at the bottom y2 - y1 = x - (x^2 - 3x + 3) x - x^2 + 3x - 3 -x^2 + 4x - 3 x^2 - 4x + 3 (x - 1)(x - 3) x = 1,3 The new interval is between 1 and 3. Integrate x^2 - 4x + 3 and you get -4/3. Take its absolute value and you get 4/3. You can also integrate -x^2 + 4x - 3. You will get 4/3
Let R be the region enclosed by the graph of y=x^2 and the line y=4...questions below please help?
a. the graph is somthing like ....y __|__ (...|...) ..\.|./ ...\|/_______x so to find the area we find intersections y=y .. x^2=4 ... so x=-2 or x=2 we do integral from intersection points of top function-botom one so [2,-2]int(4 - x^2) that gives 4x - x^3/3 |[2,-2] so 4(2)-(2)^3/3 - (4(-2) -(-2)^3/3) 8-8/3+8-8/3 16-16/3 =32/3 units squared b. well the graph i drew shows that they are pair functions, so two regions of equal area would b x=0. c. if we revolve it around the x axis than we can use disk method to cut disks perpendicular to the x axis and do the integral of their surface. surface of a disk is Pir^2 r would be 4-x^2 so Pi*[2,-2]int((4-x^2)^2) Pi*[2,-2]int(16-8x^2+x^4) Pi*(16x-8x^3/3+x^5/5)|[2,-2] Pi*((16(2)-8(2)^3/3+(2)^5/5)-(16(-2)-8... Pi*(32-64/3+32/5+32-64/3+32/5) So the volume is the product of the sum of the above fractions and Pi
How do I find the area enclosed by [math]2|x|+3|y|\le 6[/math]?
Many answers have been already written on this. But still thanks for the A2A!Let me try and highlight a few nice aspects of this question.Considering only the equation part of the question, i.e. [math]2|x| + 3|y| = 6[/math], we can make out that the resultant graph will be symmetrical about both the x-axis and the y-axis.How do I know this? …. replace [math]x[/math] with [math](-x)[/math]. Do you see any change in the equation? Nope… this means the graph is symmetrical about y-axis. Similarly check for [math](-y)[/math] and we can confirm that the graph is symmetrical about x-axis. So we have to bother about only the first quadrant.The Modu(lu)s operandi: [math]|x| = x, x \geq 0[/math] and [math]|x| = -x, x < 0[/math]So in quadrant [math]1[/math] as both [math]x[/math] and [math]y[/math] are positive, we will have the modulus inequality replaced with [math]2x + 3y \leq 6[/math]Any line on the coordinate plane divides the plane into two parts. The line itself satisfies the equation of the line and one part of the plane satisfies the < (less than) inequality and the other satisfies the > (greater than) inequality.Considering [math]2x + 3y = 6[/math] , the line intersects with the x-axis at 3 and y-axis at 2.Now let us choose a point from one side of the line - say [math](0, 0)[/math], substitute in the line and voila…. the inequality is satisfied :) So the origin part of the line satisfies the inequality; refer to the first point and we understand we can see a nice parallelogram around the origin.Required area = (Area on the first quadrant) x 4As said above, the line intersects the x and y axes at 3 and 2 respectively; the area on the first quadrant will be the area of the right triangle 2 units height and 3 units base.So required area [math] = \frac{1}{2} * 3 * 2 * 4 = 12[/math] sq.unitsHope this helps to solve more problems of this kind.
What is the area of the region enclosed by the two curves (y=e^x and y=2+3e^-x) and the y-axis?
Wolfram|Alpha: Making the world’s knowledge computabley=e^x and y=2+3e^-x) and the y-axisy=e^x= 3e(^-x)+2if F=e^x -->F= 3/F +2 =(3+2F)/FF^2-2F-3=0F= (2+(4+12)^.5 )/2= (2+4)/2 = 3F=3= e^x --> x=ln(3) -->y= e^(ln3)=3F= (2+(4-12)^.5 )/2= (-2)/2 = -1 rejected.Area= ∫ e^x dx from x=0 to x=ln(3) + ∫ 2+3e^-x dx from x=ln(3) to x=0Area= [ e^x] from x=0 to x=ln(3) + [2x-3e^-x] from x=ln(3) to x=0Area= [ e^x] from x=0 to x=ln(3) + [2x-3e^-x] from x=ln(3) to x=0Area= [ (e^0)-(e^ln3)] + [2ln3-3(e^-(ln3))-0+3e^0]Area= [ (1)-(3)] + [2ln3-3(1/3)-0+3(1)]Area= (-2)+(2ln3–1+3)Area= 2ln3= 2.197225Is it correct?when x=0 --> y=e^x=e^0 =1 --> (0,1)when x=0 --> y=3e^-x=2 =3(1)+2= 5 --> (0,5)we have triangle from three points (0,1) (0,5), (ln3, 3)Area of this triangle is =(1/2)*base* altitude= (1/2)*4*ln3=2*ln(3)so it is our answer.
What is the area enclosed by the parabola y2 = 2x and the line y = 2x?
First we solve the 2 equations to find where the parabola and line intersecty^2= 2x=yOn solving we get y=0,x=0 and y=1,x=1/2The picture is like thisThe shaded region is now the required areaSo by using integral calculus of areasIntegrate √(2x) - 2x putting limits 0 to 0.5Integration of √ax is √a (x^1.5)/1.5Integration of ax is ax^2/2By substituting limits we get answer as 1/3–1/4= 1/12Answer:1/12Note:I know the answer is not very clear because I find it difficult to type mathematical calculations but I do hope you will catch the central point of the solution and solve itThis answer is based on integration so you need to know integration
Find the total area enclosed by the curves y=abs(x) and y= x^2-2
Due to symmetry it is sufficient to do: half and multiply by 2 So we can drop the abs altogether and use y = x in the first quadrant Find point of intersection x² - 2 = x x² - x - 2 = 0 (x + 1)(x - 2) = 0 x = 2 (it's not the negtaive one hbecause it's in quadrant 1) 2∫x - (x² - 2) dx from 0 to 2 = 2∫x - x² + 2 dx from 0 to 2 = 2 [ x² /2 - x³ / 3 + 2x ] from 0 to 2 = 2 [((2)² /2 - (2)³ / 3 + 2(2) ) - ((0)² /2 - (0)³ / 3 + 2(0) )] = 2 (2 - 8/3 + 4) = 20/3 units²
Determine the area of the region bounded by the graphs of y = x^2 - 4x and y = x-4?
Lets start by graphing both functions and from there decide how we are gonna find the area bounded by the curves: Please go to this link: http://s472.photobucket.com/albums/rr82/RicardoCisneros/?action=view¤t=AREA2.jpg FRsRT OF ALL lets see that the straight line is on top and the parabola at the bottom, therefore the area can be found by AREA= INTEGRAL from 1 to 4 [ (x-4) -(x^2 -4x) ] dx Here is important to know how to find the limirts of integration you see in the graph, which are the interception of the 2 curves, we do this making them equal and solving for x, like this: x^2 - 4x = x-4 x^2 -5x +4 =0........factoring ( x - 4 ) ( x - 1 ) = 0 or x= 4 and x= 1 So what we do now is finish integrating AREA= INTEGRAL from 1 to 4 [ (x-4 -x^2 +4x ] dx AREA= INTEGRAL from 1 to 4 [ -4 +5x-x^2 ] dx INTEGRATING, POWER RULE: = [ -4x + 5x^2/2 - x^3/3 ] from 1 to 4 Please do the calculations on your own, I'll write the total area. AREA = 5/4 squared units ANSWER Questions, e-mail me