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Find the maximum area of a rectangle that can be inscribed in the ellipse . plz. explain in simple way.

^{2}+ y^{2 }<=2ax, y^{2}>=ax, x,y >= 0}Sketch the region bounded by the curves y=root of 5-x

^{2}and y=|x-1| and find its areadraw a rough sketch of the curve y = sqrt(3x+4) and find the area under the curve, above x-axis and between x=0 and x=4

Using integration find the area of that part of circle x

^{2}+y^{2}=16, which is exterior to the parabola y^{2}=6xthe locus of the foot of perpendicular drawn from the centre of the ellipseX

^{2}+ 3y^{2}=6 on any tangent to it is^{2}=4a^{2}(x-3)and the lines x=3,y=4a. [NCERT Examplar]^{}find the area bounded by y

^{2}= 4x and x^{2}= 4y^{2}=ax and x^{2}+y^{2}=4axspecific doubt: i am getting points of intersection of 2 curves as x=0 and x=3a , it is looking wrong

what is the integration of xcotx

Draw a rough sketch of the region: y

^{2}= 6ax and x^{2}+y^{2}= 16a^{2}and find the enclosed area using method of integration.find the area bounded by parabolas y

^{2}=4ax and x^{2}=4ay.find the area of the region included between the parabola y2 = x and the line x + y =2

find the area of the region enclosed between the two circles x^2+y^2=4 and (x-2)^2+y^2=4

Find the area common to the circle x

^{2}+y^{2}=16a^{2}and the parabola y^{2}=6ax.^{2 }and y-axis ?prove that the curves y2 = 4x andd x2 = 4y divide the area of the square bounded by x=0 x=4 and y = 4 and y = 0 in three equal parts?

Find the area of the region{(x,y): y^2 is less than equal to 6 ax and x^2 + y^2 is lessor equal to 16 a^2 using method of integration

using integration find area of triangle with sides x-3y+3=0,x+2y-2=0,3x+y-11=0

The area of the circle

x^{2}+y^{2}= 16 exterior to the parabolay^{2}= 6xisA.B.C.D.calculate as a limit of sum limit 1 to 2 (3x2-1)

y= -|x-1|+1

The slope of the line touching both the

parabolas y2 = 4x and x2 = 32y isfind the ratio of the areas into which the curves y

^{2}=6x divides the region bounded by x^{2}+y^{2}=16Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).

1) Find the point P on the parabola y^2 = 4ax such that area parabola, the X-axis and the tangent at P is equal to that of bounded by the parabola, the X-axis and the normal at P.

2) Find the area of the region bounded by C: y = tanx , tangent drawn at C at x = pie/4 and the x- axis.

x+2y=2

y-x=1 and

2x+y=7

find the area of a triangle whose vertices are a(1,3)b(2,5)c(3,4).

Find the area of the region bounded by the curve

y^{2}= 4xand the linex= 3Find the area of the smaller part of the circle

x^{2}+y^{2}=a^{2}cut off by the lineusing the method of integration.find the area of the region bounded by the lines 2x+y=4, 3x-2y=6 and x-3y+5=0

y= mod(x+1) +1

x= -2

x = 3

y = 0

area bounded by curve y=6x-x2 nd y=x2-2x?

Find the area of the circle 4

x^{2}+ 4y^{2}= 9 which is interior to the parabolax^{2}= 4y{(x,y): x

^{2 }+ y^{2 }^{2 }and x^{2}+y^{2}= 2a^{2}touch each otherFind the area of the region in the 1

^{st}quadrant enclosed by the x-axis, the line x=3^{1/2}y and the circle X^{2}+ Y^{2}= 4Find the area of that part of circle x

^{2 }+y^{2 }=16 which is exterior to the parabola y^{2 }=6xFind the area bounded by the curve y = 2cosx and the x-axis fromx = 0 to x = 2.

Calculate the area made by min(|x|,|y|) and max(|x|,|y|).

PLease draw the graph also.

FInd the surface area of one arch of the cycloid

x=a(t-sint) ; y=a(1-cost) ;

''0'' less than or equal to ''t'' less than or equal to ''2π''find the surface area generated by revolving the curve about x-axis(it's base)^{2}x)and find the areabetween the x axis, the curve and the ordinates

The answer: 4 square units.

The link was provided by you relating to this question, but the student could not solve the question. So, please be kind enough to solve the question .

^{2}=4a^{2}(x-1)and the lines x=1and y=4ausing integration find the area of the region enclosed between the circles x

^{2 }+ y^{2}= 1 and (x-1)^{2}+ y^{2}=1find the area of the parabola y

^{2}= 4ax bounded by its latus rectumplease explain this question and tell me.. how do we know where to shade i.e..above or below the parabola??find the area of the circle x2+y2=a2 by integration.

integration of x2+x+1/x2-x+1dx

x2=xsquareProve Sinx/n =6.

Find the area enclosed by the parabola 4

y= 3x^{2}and the line 2y= 3x+ 12