Area Under Simple Curves: Formula, Methods, Solved Examples

Area Under Simple Curves: There are formulas for finding the areas of figures such as quadrilaterals, polygons, and circles, but no specific formula exists for calculating the area under a curve. Knowing the equation of the curve, the border of the curve, and the axis containing the curve helps us to calculate the area under the curve.

Different methods are used to determine the area under simple curves, with the anti-derivative method being the most prevalent. The integration method aids in the solution of the problem and the determination of the required area. In this article, we will provide detailed information about the area under simple curves.

Formula for Area Under the Curve

As boundaries of the curve is given, the area can be determined with regard to the axes. It is possible to calculate the area under the curve in two ways: 

• with respect to the \(x\) −axis
• with respect to the \(y\) −axis

The curve is present below the axes in some cases, and partly below the axes in others. In any of these cases, we may use the derived formula to calculate the area under the curve.

Area as a Definite Integral

Let \(AD\) be the curve \(y=f(x)\) between the ordinates \(BA(x = a)\) and \(CD(x = b)\). Then, the required area is the area of region \(ABCD\).

Let \(P(x, y)\) be any point on the curve and \(Q\left( {x + \delta x,\,y + \delta y} \right)\) be a neighbouring point on the curve. Draw ordinates \(PL\) and \(QM\).

Here,

\(PL = MN = y\)

\(QN = \delta y\)

\(QM = y + \delta y\)

\(LM = \delta x\)

Let,

area \(BLPA = A\)

area \(LMQP = \delta A\)

Since \(LMNP\) is a rectangle, area \(LMNP = y\delta x\)

Similarly, 

area \(LMQR = y + \delta y\delta x\)

Clearly, area \(LMNP \le \) area \(LMQP \le \) area \(LMQR\)

\( \Rightarrow y\delta x \le \delta A \le (y + \delta y)\delta x\)

\( \Rightarrow y \le \frac{{\delta A}}{{\delta x}} \le y + \delta y\)

\(\Rightarrow y \le \mathop {\lim }\limits_{\delta x \to 0} \frac{{\delta A}}{{\delta x}} \le \mathop {\lim }\limits_{\delta x \to 0} (y + \delta y)\)

\( \Rightarrow y \le \frac{{dA}}{{dx}} \le y\)

\(\therefore \,\frac{{dA}}{{dx}} = y\)

\( \Rightarrow \int_a^b {\frac{{dA}}{{dx}}} dx = \int_a^b y dx\)

\( \Rightarrow [A]_{x = a}^{x = b} = \int_a^b y dx\)

\( \Rightarrow ( Area A\) when x = b ) – (Area A) \(x = a) = \int_a^b y dx\)

When \(x=a,PL\) coincides with \(AB\) and area \(ABLP = 0\),

\({\mathop{\rm Area}\nolimits} (ABCD) – 0 = \int_a^b y dx\)

\( \Rightarrow {\mathop{\rm Area}\nolimits} (ABCD) = \int_a^b f (x)dx\)

Hence, if \(f(x)\) is a continuous function defined on \([a, b]\), the area bounded by the curve \(y = f(x)\), the \(x\)−axis and the ordinates \(x = a\) and \(x = b\) is given by, \(\int_a^b f (x)dx\).

Note:

1. If the curve \(y = f(x)\) lies below \(x\)−axis, then the area bounded by the curve \(y=fx\), the \(x\)−axis and the ordinates \(x = a\) and \(x = b\) is negative. So, area is given by \(\int_a^b | f(x)|dx\).
2. The area bounded by the curve \(x=f(y)\), the \(y\)−axis and the abscissae \(y = c\) and \(y = d\) is given by \(\int_c^d | f(y)|dy\).
   

Algorithm to Find the Area of the Region Bounded by the Curve 

Here, the curve is \(y = f\left( x \right),{\rm{ }}x\)−axis and the ordinates area \(x = a\) and \(x = b\)

1. Make a sketch of the curve and identify the region whose area is to be found.
2. Slice the region into vertical strips. 
3. Take an arbitrary point \(P(x, y)\) on the curve and construct a representative strip of width \(dx\) having two ends of its base on \(x\)−axis at points \(\left( {x – \frac{{dx}}{2},\,0} \right)\) and \(\left( {x + \frac{{dx}}{2},\,0} \right)\) and \(\left( {x,\,0} \right)\) as the mid-point of its base.
4. Construct an approximating rectangle whose base is the same as that of the representative strip and height equal to \(|y| = |f(x)|\)
5. Find the area of the approximating rectangle as \(|y|dx = |f(x)dx|\).
6. Find the values of \(x,\,x = a\) and \(x = b\) within which the approximating rectangle can move horizontally in the given region and form the integral \(\int_a^b | f(x)|dx\).
7. Evaluate the integral obtained in the previous step. 

The value of integral so obtained is the required area.

Note:

1. If the curve \(y = f(x)\) lies above \(x\)−axis on interval \([a, b]\)  then the area of the region bounded by the curve \(y = f\left( x \right),{\rm{ }}x\)−axis and the ordinates \(x = a\) and \(x = b\) is given by

\(\int_a^b | f(x)|dx = \int_a^b f (x)dx\)

2. If the curve \(y = f(x)\) lies below \(x\)−axis on interval \([a, b]\)  then the area of the region bounded by the curve \(y = f\left( x \right),{\rm{ }}x\)−axis and the ordinates \(x =a\) and \(x = b\) is given by

\(\int_a^b | f(x)|dx = – \int_a^b f (x)dx\)

3. If \(f(x)\) is a continuous function defined on [a,b] and \(c \in (a,\,b)\) such that the curve \(y=f(x)\) lies above \(x\)−axis on \([a ,c]\) below \(x\)−axis on \([c, b]\). Then, area \(A\) of the region bounded by \(y=f(x), x\)−axis, \(x = a\) and \( x= b\) is given by

\(A = \int_a^b | f(x)|dx = \int_a^c | f(x)|dx + \int_c^b | f(x)|dx\)

Finding the Area of the Region Bounded by the Curves ; \((x=fy)\), \(y\) -axis and the lines \(y=c\) and \(y=d\)

Here, the curve \(x= f(y), y\)−axis and the lines \(y = c\) and \(y = d\)

1. Make a sketch of the curve and identify the region whose area is to be found.
2. Slice the region into horizontal strips. 
3. Take an arbitrary point \(P(x, y)\) on the curve and construct a representative strip of width \(dy\) and \(y, 0)\) as the mid-point of its base.
4. Construct an approximating rectangle whose base is same as that of the representative strip and height equal to \(|x| = |f(y)|\)
5. Find the area of the approximating rectangle as \(|x|dy = |f(y)dy|\).
6. Find the values of \(y, y = c\) and \(y = d\) within which the approximating rectangle can move horizontally in the given region and form the integral \(\int_c^d | f(y)|dy\).
7. Evaluate the integral obtained in the previous step. 

The value of integral so obtained is the required area.

Solved Examples – Area Under Simple Curves

Q.1.  Find the area enclosed by the parabola  \({y^2} = 4ax\) and its latus rectum.
Ans. Consider the following figure.

Here, the focus of the parabola is \(S(a,0)\). Hence, we need to find the area of the region OL’SL to find the required area.

Since the curve is symmetrical about \(x\)−axis, the required area\( = 2 \times {\rm{area}}\left( {{\rm{OSL}}} \right)\)

Now, we divide the area using vertical strips.

Length of the vertical strip \(= y\), and width \(=dx\)

\(\therefore \) Area of the region \(\mathrm{OLSL}=2\) area \((\mathrm{OSL})\)

\( \Rightarrow A = 2\int_0^a | y|dx\)\

\( = 2\int_0^a y dx,y \ge 0 \Rightarrow |y| = y\)

\( = 2\int_0^a {\sqrt {4ax} } dx\)

\( = 4\sqrt a \mathop \smallint \nolimits^a \sqrt x dx\)

\( = 4\sqrt a \left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^a\)

\( = 4\sqrt a \times \frac{2}{3}\left( {{a^{\frac{3}{2}}} – 0} \right)\)

\(\therefore \,A = \frac{8}{3}{a^2}\) square units

Q.2. Calculate the area of the region enclosed by the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\).
Ans. Equation of the ellipse is \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\).

We need to find the area of the region \(AB{A^\prime }{B^\prime }\).

Area enclosed by the ellipse \(=4 x\) [Area enclosed by the ellipse and the coordinate axes in one quadrant]

Length of the strip \(=y\), and width \(=dx\)

Required area is given by, \(A = 4\int_0^a y dx\)

\( = 4\int_0^a {\frac{b}{a}} \sqrt {{a^2} – {x^2}} dx\)

\( = \frac{{4b}}{a}\left[ {\frac{x}{2}\sqrt {{a^2} – {x^2}} + \frac{{{a^2}x}}{{2a}}} \right]_0^a\)

\( = \frac{{4b}}{a}\left[ {0 + \frac{{{a^2}}}{2}(1)} \right]\)

\( = \frac{{4b}}{a} \times \frac{{{a^2}}}{2} \times \frac{\pi }{2}\)

\(\therefore \,A = \pi ab\) square units

Q.3. Calculate the area of the region enclosed by \(y = – 1,\,y = 2,\,x = {y^3}\)  and \(x = 0\).
Ans.

The graph of the curves \(y = – 1,\,y = 2\), and \(x = {y^3}\) is given as:

The shaded region in the above graph is the required area.

Dividing the area with horizontal strips, we observe that the approximating rectangle has length \(= x\), and width \(= dy\)

And, area \( = |x|dy\)

Since the approximating rectangle can move vertically from \(y = −1\) to \(y = 2\).  So, required area \(A\) is given by

\(A = \int_{ – 1}^2 | x|dy\)

\( = \int_{ – 1}^0 | x|dy + \int_0^2 | x|dy\)

\( = \int_{ – 1}^0 {\left( { – x\mid dy + \int_0^2 x dy} \right.} \)

\( = – \int_{ – 1}^0 x dy + \int_0^2 x dy\)

\(A = \int_{ – 1}^0 – {y^3}dy + \int_0^2 {{y^3}} dy\)

\(A = – \left[ {\frac{{{y^4}}}{4}} \right]_{ – 1}^0 + \left[ {\frac{{{y^4}}}{4}} \right]_0^2\)

\( = \frac{1}{4} + 4\)

\(\therefore \,A = \frac{{17}}{4}\) square units

Q.4. Find the area bounded by the curve \(y = x|x|,\,x\)−axis and the ordinates \(x = – 3\) and \(x = 3\).
Ans.

Given equation of the curve is

\(y = x|x|\)

\(\Rightarrow y = x|x| = \left\{ {{x^2},x \ge 0 – {x^2},x < 0} \right.\)

The graph of \(y = x|x|\) is given below, and the region bounded by \(y = x|x|,\,x\)−axis, and the ordinates \(x = −3\), and \(x = 3\) is shaded.

The area of the shaded region is twice the area of the shaded region in the first quadrant.

Let us divide the region in first quadrant into vertical strips. 

The approximating rectangle shown in the above figure has length \( = \left| {{y_1}} \right|\). As it can move between \(x = 0\) and \(x = 3\). 

Therefore,

Required area \( = 2\int_0^3 {\left| {{y_1}} \right|} dx = 2\int_0^3 {{y_1}} dx\left[ {\,{y_1} \ge 0\,\therefore \,\left| {{y_1}} \right| = {y_1}} \right]\)

\(A = 2\int_0^3 {{x^2}} dx\)

\( = 2\left[ {\frac{{{x^3}}}{3}} \right]_0^3\)

\(\therefore \,A = 18\) sq. units. 

Q.5. Using the method of integration find the area-bounded by the curve \(|x| + |y| = 1\).
Ans.
Given:
\(|x| + |y| = 1\)
\( \Rightarrow x + y = 1,\,{\rm{when }} x \ge 0,\,y \ge 0\)
\( \Rightarrow – x + y = 1,\,{\rm{when }} x < 0,\,y \ge 0\)
\( \Rightarrow x – y = 1,{\rm{ when }} x \ge 0,\,y < 0\)
\( \Rightarrow – x – y = 1,{\rm{ when } }x < 0,\,y < 0\)

Thus, the graph of the given curve is as follows:

Area, \(A = {\rm{Area }} COBO + {\rm{Area }} OABO + {\rm{Area }} COBC + {\rm{Area }} ODAO\)

\(A = \int_{ – 1}^0 {\left| {{y_1}} \right|} dx + \int_0^1 {\left| {{y_2}} \right|} dx + \int_{ – 1}^0 {\left| {{y_3}} \right|} dx + \int_0^1 {\left| {{y_4}} \right|} dx\)

\(A = \int_{ – 1}^0 | 1 + x|dx + \int_0^1 | 1 – x|dx + \int_{ – 1}^0 | – x – 1|dx + \int_0^1 | x – 1|dx\)

\(A = \int_{ – 1}^0 {(1 + x)} dx + \int_0^1 {(1 – x)} dx + \int_{ – 1}^0 {(x + 1)} dx + \int_0^1 {(1 – x)} dx\)

\(A = 2\int_{ – 1}^0 {(x + 1)} dx + 2\int_0^1 {(1 – x)} dx = 2\left[ {\frac{{{x^2}}}{2} + x} \right]_{ – 1}^0 + 2\left[ {x – \frac{{{x^2}}}{2}} \right]_0^1\)

\(A = 2\left\{ {0 – \left( {\frac{1}{2} – 1} \right)} \right\} + 2\left\{ {\left( {1 – \frac{1}{2}} \right) – 0} \right\}\)

\({\rm{ = 1 + 1}}\)

\(\therefore \,A = 2\) sq. units

Summary

In this article, the formulae to find the area under the curves is discussed in detail. It also lists the algorithm to find the area of the region bounded by the curve \(y = f(x), x\)−axis and the ordinates \(x = a\) and \(x = b\), the algorithm to find the area of the region bounded by the curve \(x = f(y), y\)−axis and the lines \(y = c\) and \(y = d\).

Starting with plotting the curve, we can then proceed to break the region into strips. After calculating the area of one of the strips, it is then integrated to find the complete area. The solved examples demonstrate how these formulas can be applied to find the area under simple curves.

Frequently Asked Questions (FAQs)

Q.1. How do you find the area under a simple curve?
Ans. The area under a curve between two points is calculated using a definite integral between those two points. To get the area under the curve \(y = f(x)\) between \(x = a\) and \(x = b\), integrate \(y = f(x)\) between the limits of \(a\) and \(b\).
Area under the curve \( = \int_a^b f (x)dx\)

Q.2. What does “area under the curve” mean?
Ans. The region circumscribed by the function, \(f(x)\), the vertical lines denoting the function’s boundaries on both the sides and the \(x\)−axis is known as the area under the curve.

The graph above depicts the continuous function’s area under the curve. The function’s vertical boundaries are represented by the interval \([a, b]\).

Q.3. What is the first step toward finding the area between two curves?
Ans. The first step for finding the area between two curves is to plot the two curves on the same graph. Then we need to find the coordinates of the points of intersection of the two curves to find out the boundaries of the region whose area has to be calculated. Further, apply the formula for finding the area between two curves.

Q.4. Is the area under a curve always positive?
Ans. An integral symbolizes the area between the curve of a function and the \(x\)−axis. If the function lies below the x−axis, then we get the negative value of the integral. For the curve \(y = f(x)\) that lies below \(x\)−axis, the area bounded by the curve, the \(x\)−axis and the ordinates \(x = a\) and \(x = b\) is given by:

\(A = \int_a^b | f(x)|dx\)

Since we are taking an absolute value of the function, the area under the curve will never be negative. It is always positive.

Q.5. What is the formula for calculating the area of two curves?
Ans:
When two curves, \(y = f(x)\) and \(y = g(x)\), intersect at two points, with \(x\)−coordinates as \(x =a\) and \(x =b\), such that:

\(f(x) \ge g(x)\), for \(a \le x \le b\)

\( \Rightarrow y = f(x)\) is above \(y = g(x)\) over the interval \(a \le x \le b\)

The area enclosed by these two curves can be calculated using the formula:

Enclosed Area \( = \int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]} dx\).

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