Area under the curve is calculated by different methods, of i beg your pardon the antiderivative technique of detect the area is most popular. The area under the curve deserve to be uncovered by learning the equation of the curve, the limits of the curve, and the axis enclosing the curve. Generally, we have formulas because that finding the locations of constant figures such as square, rectangle, quadrilateral, polygon, circle, however there is no identified formula to uncover the area under the curve. The procedure of integration help to solve the equation and also find the forced area.

You are watching: What does the area under the curve represent

For recognize the areas of irregular airplane surfaces the techniques of antiderivatives are an extremely helpful. Here we shall learn just how to find the area under the curve with respect to 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 discover Area Under The Curve 3 Formula for 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 top top Area Under The Curve

## How to uncover Area Under The Curve?

The area under the curve deserve to be calculated with three simple steps. First, we need to know the equation the the curve(y = f(x)), the limits across which the area is to it is in calculated, and the axis enclosing the area. Secondly, we have to find the integration (antiderivative) of the curve. Finally, we require to apply the top limit and lower border to the integral answer and also take the distinction to obtain the area under the curve.

Area = $$_a\int^b y.dx$$

= $$_a\int^b f(x).dx$$

=$$^b_a$$

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

## Different methods to find Area Under The Curve

The area under the curve can be computed using 3 methods. Also, the technique used to uncover the area under the curve relies on the need and the obtainable data inputs, to uncover the area under the curve. Here we chandelier look into the listed below three techniques to find the area under the curve.

Method - I: here the area under the curve is damaged down right into the smallest possible rectangles. The summation of the area of these rectangles offers the area under the curve. For a curve y = f(x), it is broken into numerous rectangles the width$$\delta x$$. Below we border the number of rectangles as much as infinity. The formula because that the total area under the curve is A =$$\lim_x \rightarrow \infty\sum _i = 1^nf(x).\delta x$$.

Method - II: This an approach also provides asimilar procedure as the above to discover the area under the curve. Here the area under the curve is separated into a few rectangles. Further, the areas of this rectangles are included to acquire the area under the curve. This method is one easy method to find the area under the curve, yet it only gives an approximate value of the area under the curve.

Method - III: This technique makes usage of the integration procedure to find the area under the curve. To discover the area under the curve through this an approach integration we require the equation that the curve, the knowledge of the bounding lines or axis, and also the border limiting points. For a curve having actually an equation y = f(x), and also bounded by the x-axis and with limit worths of a and also b respectively, the formula for the area under the curve is A = $$_a\int^b f(x).dx$$

## Formula for Area Under the Curve

The area of the curve can be calculated v respect come the various axes, as the border for the given curve. The area under the curve have the right to be calculated v respect to the x-axis or y-axis. For one-of-a-kind cases, the curve is listed below the axes, and also partly below the axes. For all these instances we have actually the acquired formula to uncover the area under the curve.

Area with respect to the x-axis: here we shall first look in ~ the area attached by the curve y = f(x) and the x-axis. The below figures gift the area fastened by the curve and the x-axis. The bounding values for the curve v respect to the x-axis room a and also b respectively. The formula to discover the area under the curve v respect come the x-axis is A = $$_a\int^b f(x).dx$$ Area with respect to the y-axis: The area of the curve bounded through the curve x = f(y), the y-axis, across the present y = a and y = b is provided by the following below expression. Further, the area in between the curve and the y-axis deserve to be understood from the below graph.

A = $$_a\int ^bx.dy = _a\int^b f(y).dy$$ Area below the axis: The area that the curve below the axis is a an unfavorable value and hence the modulus that the area is taken. The area the the curve y = f(x) listed below the x-axis and bounded through the x-axis is obtained by taking the borders a and b. The formula because that the area above the curve and the x-axis is together follows.

A = |$$_a\int ^bf(x).dx$$| Area above and listed below the axis: The area of the curve i m sorry is partly below the axis and also partly over the axis is separated into 2 areas and also separately calculated. The area under the axis is negative, and hence a modulus of the area is taken. As such the overall area is equal to the sum of the two areas($$A = |A_1 |+ A_2$$).

A =|$$_a\int ^bf(x).dx$$| +$$_b\int ^cf(x).dx$$ ## Area Under TheCurve - Circle

The area that the circle is calculation by an initial calculating the area the the component of thecircle in the an initial quadrant. Below the equation the the circle x2+ y2= a2is changed to an equation that a curve as y =√(a2 - x2). This equation the the curve is provided to discover the area through respect to the x-axis and also the boundaries from 0 come a. The area that the circle is 4 times the area of the quadrant the the circle. The area the the quadrant is calculated by completely the equation that the curve across the limits in the very first quadrant.

A = 4$$\int^a_0 y.dx$$

= 4$$\int^a_0 \sqrta^2 - x^2.dx$$

= 4$$<\fracx2\sqrta^2 - x^2 + \fraca^22Sin^-1\fracxa>^a_0$$

= 4<((a/2)× 0 + (a2/2)Sin-11) - 0>

= 4(a2/2)(π/2)

=2πr

Hence the area the the one isπa2square units.

## Area Under a Curve - Parabola

A parabola has actually an axis the divides the parabola right into two symmetric parts. Right here we take a parabola the is symmetric follow me the x-axis and has an equation y2= 4ax. This deserve to be revolutionized as y =√(4ax). We an initial find the area the the parabola in the first quadrant through respect to the x-axis and also along the limits from 0 come a. Below we incorporate the equation within the border and double it, to achieve the area the the entirety parabola. The derivations because that the area of the parabola is together follows.

\beginalignA &=2 \int_0^a\sqrt4ax.dx\\ &=4\sqrt a \int_0^a\sqrt x.dx\\& =4\sqrt a<\frac23.x^\frac32>_0^a\\&=4\sqrt a ((\frac23.a^\frac32) - 0)\\&=\frac8a^23\endalign

Therefore the area underthe curve enclosed by the parabola is $$\frac8a^23$$ square units.

## Area Under a Curve - Ellipse

The equation that the ellipse with the major axis of 2a and also a minor axis the 2bis x2/a2+ y2/b2= 1. This equation deserve to be reinvented in the type as y = b/a .√(a2- x2). Below we calculation the area bounded by the ellipse in the first coordinate and also with the x-axis, and also further main point it with 4 to attain the area the the ellipse. The boundary boundaries taken on the x-axis is from 0 come a. The calculations for the area of the ellipse room as follows.

\beginalignA &=4\int_0^a y.dx \\&=4\int_0^4 \fracba.a^2 - x^2.dx\\&=\frac4ba<\fracx2.\sqrta^2 - x^2 + \fraca^22Sin^-1\fracxa>_0^a\\&=\frac4ba<(\fraca2 \times 0) + \fraca^22.Sin^-11) - 0>\\&=\frac4ba.\fraca^22.\frac\pi2\\&=\pi ab\endalign

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 also a linecan be conveniently calculate by taking the distinction of the locations of one curve andthe area under the line. Below the boundary v respect to the axis for both the curve and the lineis the same. The listed below figure shows thecurve$$y_1$$ = f(x), and also the heat $$y_2$$ = g(x), and the objective is to discover the area between the curve and also the line. Right here we take the integral that the distinction of the 2 curves and also apply the boundariesto uncover the result area.

A =$$\int^b_a .dx$$

## Area Under a Curve - in between Two Curves

The area in between two curves deserve to be conveniently calculated by acquisition the distinction of the areas of one curve native the area of another curve. Right here the boundary v respect come the axis for both the curve is the same. The below figure reflects two curve $$y_1$$ = f(x), and $$y_2$$ = g(x), and the target is to discover the area between these two curves. Here we take the integral of the distinction of the two curves and apply the boundariesto find the resultant.

A =$$\int^b_a .dx$$ Example 1: uncover the area under the curve, for the an ar bounded through the circle x2+ y2 = 16in the first quadrant.

See more: Why Did The Boy Have 5 Loaves And 2 Fish ? Feeding The Multitude

Solution:

The offered equation that the one isx2+ y2 = 16

Simplifying this equation we have actually y = $$\sqrt4^2 - x^2$$

A = $$\int^4_0 y.dx$$

= $$\int^4_0 \sqrt4^2 - x^2.dx$$

= $$<\fracx2\sqrt4^2 - x^2 + \frac4^22Sin^-1\fracx4>^4_0$$

= <((4/2)× 0 + (16/2)Sin-11) - 0>

= (16/2)(π/2)

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

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

Solution:

The offered equation that the ellipse is.x2/36 + y2/25= 1

This deserve to be reinvented to attain y = $$\frac56\sqrt6^2 - x^2$$

\beginalignA &=4\int_0^6 y.dx \\&=4\int_0^6 \frac56.\sqrt6^2 - x^2.dx\\&=\frac206<\fracx2.\sqrt6^2 - x^2 + \frac6^22Sin^-1\fracx6>_0^6\\&=\frac206<(\frac62 \times 0) + \frac6^22.Sin^-11) - 0>\\&=\frac206.\frac362.\frac\pi2\\&=30\pi \endalign