Area under the curve is calculation by various methods, of i beg your pardon the antiderivative an approach of finding the area is most popular. The area under the curve have the right to be found by knowing the equation that the curve, the borders of the curve, and the axis enclosing the curve. Generally, we have actually formulas because that finding the areas of continuous figures such as square, rectangle, quadrilateral, polygon, circle, yet there is no characterized formula to discover the area under the curve. The process of integration help to solve the equation and find the required area.

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For detect the areas of irregular airplane surfaces the methods of antiderivatives are very helpful. Here we chandelier learn how to discover the area under the curve v respect come the axis, to find the area in between a curve and also a line, and to discover the area in between two curves.

 1 How to discover Area Under The Curve? 2 Different techniques to find Area Under The Curve 3 Formula because that Area Under The Curve 4 Area Under The Curve - Circle 5 Area Under The Curve - Parabola 6 Area Under The Curve - Ellipse 7 Area between a Curve and A-Line 8 Area in between Two Curves 9 Solved Examples 10 Practice Questions 11 FAQs ~ above Area Under The Curve

How to find Area Under The Curve?

The area under the curve deserve to be calculated v three basic steps. First, we require to know the equation the the curve(y = f(x)), the limits throughout which the area is to be calculated, and also the axis enclosing the area. Secondly, we have to uncover the integration (antiderivative) that the curve. Finally, we need to use the top limit and lower border to the integral answer and also take the difference to acquire the area under the curve.

Area = (_aint^b y.dx )

= (_aint^b f(x).dx)

=( ^b_a)

=( g(b) - g(a))

Different techniques to discover Area Under The Curve

The area under the curve have the right to be computed using three methods. Also, the method used to uncover the area under the curve counts on the need and also the obtainable data inputs, to uncover the area under the curve. Below we shall look into the below three techniques to find the area under the curve.

Method - I: right here the area under the curve is broken down right into the smallest possible rectangles. The summation that the area of these rectangles offers the area under the curve. Because that a curve y = f(x), that is damaged into countless rectangles the width(delta x). Here we border the variety of rectangles approximately infinity. The formula because that the complete area under the curve is A =(lim_x ightarrow inftysum _i = 1^nf(x).delta x).

Method - II: This method also uses asimilar procedure together the above to discover the area under the curve. Right here the area under the curve is divided into a couple of rectangles. Further, the locations of these rectangles are added to acquire the area under the curve. This method is one easy technique to discover the area under the curve, however it only provides an approximate value of the area under the curve.

Method - III: This technique makes use of the integration procedure to uncover the area under the curve. To find the area under the curve through this method integration we require the equation that the curve, the knowledge of the bounding currently or axis, and the border limiting points. For a curve having actually an equation y = f(x), and bounded by the x-axis and also with limit worths of a and b respectively, the formula because that the area under the curve is A = ( _aint^b f(x).dx)

Formula because that Area Under the Curve

The area that the curve can be calculated with respect come the various axes, as the border for the provided curve. The area under the curve can be calculated v respect come the x-axis or y-axis. For special cases, the curve is below the axes, and also partly listed below the axes. For all these instances we have the derived formula to uncover the area under the curve.

Area with respect come the x-axis: right here we shall first look in ~ the area attached by the curve y = f(x) and also the x-axis. The listed below figures presents the area enclosed by the curve and also the x-axis. The bounding values for the curve through respect come the x-axis are a and b respectively. The formula to find the area under the curve v respect come the x-axis is A = (_aint^b f(x).dx)

Area v respect come the y-axis: The area that the curve bounded by the curve x = f(y), the y-axis, across the currently y = a and also y = b is provided by the following below expression. Further, the area between the curve and also the y-axis have the right to be interpreted from the below graph.

A = (_aint ^bx.dy = _aint^b f(y).dy)

Area below the axis: The area that the curve below the axis is a an adverse value and also hence the modulus that the area is taken. The area of the curve y = f(x) listed below the x-axis and also bounded by the x-axis is derived by acquisition the limits a and also b. The formula because that the area above the curve and the x-axis is as follows.

A = |(_aint ^bf(x).dx)|

Area over and listed below the axis: The area the the curve i beg your pardon is partly listed below the axis and partly over the axis is divided into 2 areas and also separately calculated. The area under the axis is negative, and also hence a modulus that the area is taken. Therefore the in its entirety area is same to the amount of the 2 areas((A = |A_1 |+ A_2)).

A =|(_aint ^bf(x).dx)| +(_bint ^cf(x).dx)

Area Under TheCurve - Circle

The area the the circle is calculate by first calculating the area the the component of thecircle in the an initial quadrant. Here the equation the the one x2+ y2= a2is adjusted to one equation of a curve together y =√(a2 - x2). This equation that the curve is offered to find the area through respect come the x-axis and the limits from 0 to a.

The area that the circle is 4 times the area the the quadrant of the circle. The area that the quadrant is calculated by completely the equation the the curve throughout the borders in the first quadrant.

A = 4(int^a_0 y.dx)

= 4(int^a_0 sqrta^2 - x^2.dx)

= 4(^a_0)

= 4<((a/2)× 0 + (a2/2)Sin-11) - 0>

= 4(a2/2)(π/2)

=2πr

Hence the area the the circle isπa2square units.

Area Under a Curve - Parabola

A parabola has an axis the divides the parabola into two symmetric parts. Below we take a parabola the is symmetric follow me the x-axis and also has an equation y2= 4ax. This have the right to be transformed as y =√(4ax). We an initial find the area of the parabola in the first quadrant with respect to the x-axis and along the borders from 0 come a. Below we integrate the equation in ~ the boundary and double it, to obtain the area the the totality parabola. The derivations because that the area that the parabola is as follows.

(eginalignA &=2 int_0^asqrt4ax.dx\ &=4sqrt a int_0^asqrt x.dx\& =4sqrt a_0^a\&=4sqrt a ((frac23.a^frac32) - 0)\&=frac8a^23endalign)

Therefore the area underthe curve attached by the parabola is (frac8a^23) square units.

Area Under a Curve - Ellipse

The equation that the ellipse v the major axis that 2a and a boy axis that 2bis x2/a2+ y2/b2= 1. This equation can be reinvented in the kind as y = b/a .√(a2- x2). Right here we calculate the area bounded by the ellipse in the first coordinate and also with the x-axis, and also further multiply it through 4 to attain the area the the ellipse. The boundary boundaries taken top top the x-axis is from 0 come a. The calculations for the area of the ellipse are as follows.

(eginalignA &=4int_0^a y.dx \&=4int_0^4 fracba.a^2 - x^2.dx\&=frac4ba_0^a\&=frac4ba<(fraca2 imes 0) + fraca^22.Sin^-11) - 0>\&=frac4ba.fraca^22.fracpi2\&=pi abendalign)

Therefore the area that the ellipse isπab sq units.

Area Under The Curve - between a Curve and A-Line

The area in between a curve and a linecan be conveniently calculation by acquisition the distinction of the areas of one curve andthe area under the line. Right here the boundary v respect come the axis for both the curve and the lineis the same. The below figure reflects thecurve(y_1) = f(x), and also the heat (y_2) = g(x), and the objective is to discover the area in between the curve and the line. Here we take the integral of the distinction of the 2 curves and also apply the boundariesto discover the resultant area.

A =(int^b_a .dx)

Area Under a Curve - between Two Curves

The area between two curves deserve to be conveniently calculated by acquisition the distinction of the areas of one curve from the area of one more curve. Below the boundary through respect come the axis for both the curve is the same. The listed below figure reflects two curves (y_1) = f(x), and (y_2) = g(x), and the objective is to discover the area between these two curves. Right here we take the integral that the distinction of the 2 curves and also apply the boundariesto uncover the resultant.

A =(int^b_a .dx)

Example 1: find the area under the curve, for the region bounded by the one x2+ y2 = 16in the first quadrant.

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Solution:

The provided equation that the circle isx2+ y2 = 16

Simplifying this equation we have y = (sqrt4^2 - x^2)

A = (int^4_0 y.dx)

= (int^4_0 sqrt4^2 - x^2.dx)

= (^4_0)

= <((4/2)× 0 + (16/2)Sin-11) - 0>

= (16/2)(π/2)

=4πAnswer: therefore the area the the region bounded through the one in the first quadrant is4π sq units

Example 2: discover the area under the curve, forthe an ar enclosed through the ellipsex2/36 + y2/25= 1.

Solution:

The given equation of the ellipse is.x2/36 + y2/25= 1

This deserve to be reinvented to acquire y = (frac56sqrt6^2 - x^2)

(eginalignA &=4int_0^6 y.dx \&=4int_0^6 frac56.sqrt6^2 - x^2.dx\&=frac206_0^6\&=frac206<(frac62 imes 0) + frac6^22.Sin^-11) - 0>\&=frac206.frac362.fracpi2\&=30pi endalign)