Area between Two Curves: Distance is the integral of speed with respect to time. As a result, the area between the two curves represents how far one particle travelled in comparison to the other. The area between two curves is the space between two intersecting curves that can be determined with integral calculus.

Beginning with the simple example in which one function value is always bigger than the other, we compute the area between two curves that are functions of \(x.\) Then we look at scenarios where the function graphs cross. Finally, we will look at how to find the area between two curves that are functions of \(y.\) These problems not only provide an important application of definite integrals but also provide insight into the nature of the two curves involved.

How to Calculate Area Between Two Curves?

The area between two curves is the space between two intersecting curves that can be determined with integral calculus. When we have the equation of two curves and their intersection points, we can use integration to get the area under them. We have two functions \(f(x)\) and \(g(x)\) in the image, and we need to calculate the area between these two curves in the shaded section. The area of the shaded section may then be simply calculated using integration.

There are two methods to find the area between two curves.

Vertical Strips Method to Find Area Between Two Curves

If two curves are such that one is below the other and we wish to find the area of the region bounded by them and on the left and right by vertical lines. In such cases, we may use the following procedure.

Step 1: Draw given curves \(y=f(x)\) and \(y=g(x).\)

Step 2: Draw the vertical lines \(x=a\) and \(x=b.\)

Step 3: Identify the region included between the curves and vertical lines drawn in step \(1.\)

Step 4: Take an arbitrary point \(P\left( {x,{y_1}} \right)\) on one of the curves, say \(y=f(x),\) and draw a vertical line through \(P\) to meet the other curves \(y=g(x),\) at \(Q\left( {x,{y_2}} \right).\) Clearly, \({y_1} = f(x)\) and \({y_2} = g(x)\)

Step 5: Draw a vertical rectangle of width\(=dx\) and height (or length)\( = \left| {{y_1} – {y_2}} \right| = |f(x) – g(x)|\) such that \(P\left( {x,{y_1}} \right)\) and \(Q\left( {x,{y_2}} \right)\) are mid-points of horizontal sides \(AB\) and \(CD.\)

Step 6:  Find the area of the approximating rectangle drawn in step \(4.\) Let \(ΔA\) be the area. Then, 

\(\Delta A = |f(x) – g(x)|dx\)

Step 7: Use the formula

\(A = \int_a^b | f(x) – g(x)|dx\)

Area Between Two Compound Curves

If two function \(f(x)\) and \(g(x)\) are defined on \([a, b]\) that can be divided into sub-intervals \([a, c]\) and \([c, b]\) such that

\(f(x) \ge g(x)\) for all \(x \in [a, c]\)

\(f(x) \le g(x)\) for all \(x \in [c, b]\)

Then, the area \(A\) between the two curves on the interval \([a, b]\) is given by,

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

\( \Rightarrow A = \int_a^c {\{ f(} x) – g(x)\} dx + \int_c^b {\{ g(} x) – f(x)\} dx\)

Horizontal Strips Method to Find Area Between Two Curves

To find the area between two curves of the form \(x = \phi (y)\) and \(x = \psi (y)\) on the interval \([c,d]\) on the \(y−\)axis, we form horizontal strips. The procedure is as explained below. 

Step 1:  Draw given curves \(x = \phi (y)\) and \(x = \psi (y).\)
Step 2: Draw horizontal lines \(y=c\) and \(y=d.\)
Step 3:  Identify the region included between the curves and horizontal lines drawn in step 1.
Step 4: Take an arbitrary point \(P\left( {{x_1},y} \right)\) on one of the curves say \( x = \phi (y),\) and draw a horizontal line through \(P\) to meet the other curve \(x = \phi (y)\) at \(Q\left( {{x_2},y} \right).\)Clearly, \({x_1} = \phi (y)\) and \({x_2} = \psi (y).\)
Step 5: Draw horizontal approximating rectangle of width\(=Δy,\) length\( = \left| {{x_2} – {x_1}} \right| = |\psi (y) – \phi (y)|\) such that \(P\left( {{x_1},y} \right)\) and \(Q\left( {{x_2},y} \right)\) are mid-point of horizontal sides \(AB\) and \(CD\) as shown in the figure given below.

Step 6: Let \(∆A\) be the areaof the approximating rectangle drawn in step 4. Then,

\(\Delta A = |\psi (y) – \phi (y)|\Delta y\)

Step 7: Use the formula

\(A = \int_c^d | \psi (y) – \phi (y)|dy\)

Area Between Two Compound Curves

If two functions \(\phi (y)\) and \(\psi (y)\) are defined on \(\left[ {c,d} \right]\) on \(y−\)axis which can be divided into sub-intervals \(\left[ {c, e} \right]\) and \(\left[ {e,d} \right]\) such that \(\phi (y) \ge \psi (y)\) for all \(y \in [c, e]\) and \(\phi (y) \le \psi (y)\) for all \(y \in [e, d].\) That is the curve \(x = \phi (y)\) is on the right of the curve \(x = \psi (y)\) over the sub-interval \(\left[ {c, e} \right]\) and the curve \(x = \psi (y)\) is on the right of the curve \(x = \phi (y)\) as shown in the figure given below:

Then, the area \(A\) between the curves on the interval \(\left[ {c,d} \right]\) on \(y−\)axis is given by,

\(A = \int_0^d | \phi (y) – \psi (y)|dy = \int_c^e | \phi (y) – \psi (y)|dy + \int_e^d | \phi (y) – \psi (y)|dy\)

\( \Rightarrow A = \int_c^e {\{ \phi (y) – \psi (y)\} } dy + \int_\theta ^d {\{ \psi (y) – \phi (y)\} } dy\)

Solved Examples – Area Between Two Curves

Q.1.Find the area of the region included between the parabola \(y = \frac{{3{x^2}}}{4}\) and \(3x−2y+12=0.\)
Ans: Given equation of the curves are:
\(y = \frac{{3{x^2}}}{4}. \ldots (i)\)
\(3x – 2y + 12 = 0 \ldots .(ii)\)
The curve in equation \((i)\) represents an upward parabola with vertex at the origin.
And, \(3x−2y+12=0\) represents an equation of a line which intersects the \(x−\)axis at \((−4,0)\) and it intersects the \(y−\)axis at \((0,6).\)
To find the point of intersection of the given curves,
\(3x−2y+12=0\)
\( \Rightarrow 3x – 2 \times \frac{{3{x^2}}}{4} + 12 = 0\) [Substituting the value of \(y\) from equation \((i)\)]
\( \Rightarrow 3x – \frac{{3{x^2}}}{2} + 12 = 0\)
\( \Rightarrow – 3{x^2} + 6x + 24 = 0\)
\( \Rightarrow 3{x^2} – 6x – 24 = 0\)
\( \Rightarrow {x^2} – 2x – 8 = 0\)
\( \Rightarrow {x^2} – 4x + 2x – 8 = 0\)
\( \Rightarrow x(x – 4) + 2(x – 4) = 0\)
\( \Rightarrow (x + 2)(x – 4) = 0\)
\( \Rightarrow x = – 2\) or \(x=4\)
Since \(y = \frac{{3{x^2}}}{4}\)
\( \Rightarrow y = 3\) or \(12\)
Thus, the given curves are intersected at the points \((−2,3)\) and \((4,12).\)

Now, the area between two curves is the area of the shaded region as shown in the above figure.
Now, let us slice the shaded region into vertical strips. We find that each vertical strip runs from the parabola to the line. So, the approximating rectangle shown in the above figure has width\(=dx,\) and length\( = \left| {{y_2} – {y_1}} \right|.\) Thus, Area of the rectangular strip\( = \left| {{y_2} – {y_1}} \right|dx.\) Since the approximating rectangle can move horizontally from \(x=−2\) to \(x=4.\) So, area \(A\) of the region included between curves \((i)\) and \((ii)\) is given by,
\(A = \int_{ – 2}^4 {\left| {{y_2} – {y_1}} \right|} dx = \int_{ – 2}^4 {\left( {{y_2} – {y_1}} \right)} dx\)
Since \(\left( {x,{y_1}} \right)\) and \(\left( {x,{y_2}} \right)\) lies on equations \((i),\) and \((ii),\) we have
\({y_1} = \frac{{3{x^2}}}{4}\) and \({y_2} = \frac{{3x + 12}}{2}\)
Required Area\( = \int_{ – 2}^4 {\left( {\frac{{3x + 12}}{2} – \frac{{3{x^2}}}{4}} \right)} dx\)
\( \Rightarrow A = \left[ {\frac{3}{4}{x^2} + 6x – \frac{{{x^3}}}{4}} \right]_{ – 2}^4\)
\( = \left( {\frac{3}{4} \times 16 + 6 \times 4 – \frac{{64}}{4}} \right) – \left( {\frac{3}{4} \times 4 + 6( – 2) + \frac{8}{4}} \right)\)
\(\therefore A = 27\,{\rm{sq}}{\rm{.units}}\)

Q.2. Find the area of the region included between the parabolas \({y^2} = 4ax\) and \({x^2} = 4ay,\) where \(a>0.\)
Ans: Given equation of the curves are
\({y^2} = 4ax……\left( i \right)\)
\({x^2} = 4ay…….\left( {ii} \right)\)
(i) and (ii) represent parabolas in standard forms.
For point of intersection of the given curves, we solve equations \((i)\) and \((ii)\)
Putting \(y = \frac{{{x^2}}}{{4a}}\) from (ii) into (i), we get
\({\left( {\frac{{{x^2}}}{{4a}}} \right)^2} = 4ax\)
\( \Rightarrow {x^4} = 64{a^3}x\)
\( \Rightarrow x\left( {{x^3} – 64{a^3}} \right) = 0\)
\( \Rightarrow x = 0\) or, \(x = 4a\)
From equation \((i),\)
\(x = 0 \Rightarrow y = 0,\) and \(x = 4a \Rightarrow y = 4a\)
Thus, the point of intersection of two curves are \((0,0)\) and \((4a,4a).\)

Let us slice this region into vertical strips.
Here, width \(=dx,\) length\( = \left| {{y_2} – {y_1}} \right|\) and area \( = \left| {{y_2} – {y_1}} \right|dx.\)
Since the approximating rectangle can move between \(x=0\) and \(x=4a,\) the required area \(A\) is given by,
\(A = \int_0^{4a} {\left| {{y_2} – {y_1}} \right|} dx = \int_0^{4a} {\left( {{y_2} – {y_1}} \right)} dx\)
\(P\left( {x,{y_1}} \right)\) and \(Q\left( {x,{y_2}} \right)\) lies on the curves \({x^2} = 4ay\) and \({y^2} = 4ax\) respectively.
So, \({y_1} = \frac{{{x^2}}}{{4a}}\) and \({y_2} = \sqrt {4ax} \)
\( \Rightarrow A = \int_0^{4a} {\left\{ {2\sqrt {ax} – \frac{{{x^2}}}{{4a}}} \right\}} dx\)
=\({ = \left[ {\frac{{4\sqrt a }}{3}{x^{3/2}} – \frac{{{x^3}}}{{12a}}} \right]_0^{4a}}\)
\( = \frac{{32{a^2}}}{3} – \frac{{16{a^2}}}{3}\)
\(\therefore A = \frac{{16{a^2}}}{3}{\rm{sq}} \cdot {\rm{units}}\)

Q.3. Find the area of the smaller region bounded by the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) and the straight line \(\frac{x}{a} + \frac{y}{b} = 1.\)
Ans: The equations of the given curves are
\(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1…..\left( i \right)\)
\(\frac{x}{a} + \frac{y}{b} = 1……\left( {ii} \right)\)
Here, equation \((i)\) is an equation of the ellipse, and equation \((ii)\) is an equation of the straight line which intersects the \(x−\)axis and \(y−\)axis at point \((a,0)\) and \((0,b)\) respectively.

The shaded region represents the region between the given curves.
Let us slice this region into vertical strips.
The approximating rectangle has width\(=dx,\) and length\( = \left| {{y_2} – {y_1}} \right|,\) and area\( = \left| {{y_2} – {y_1}} \right|dx\)
Since the approximating rectangle can move from \(x=0\) to \(x=a.\) So, the required area is given by
\(A = \int_0^a {\left| {{y_2} – {y_1}} \right|} dx = \int_0^a {\left( {{y_2} – {y_1}} \right)} dx \ldots ..(iii)\)
\(P\left( {x,{y_1}} \right)\) lies on \(\frac{x}{a} + \frac{y}{b} = 1\) and point \(Q\left( {x,{y_2}} \right)\) lies on \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\)
\(\frac{x}{a} + \frac{{{y_1}}}{b} = 1\) and \(\frac{{{x^2}}}{{{a^2}}} + \frac{{y_2^2}}{{{b^2}}} = 1\)
\( \Rightarrow {y_1} = \frac{b}{a}(a – x)\)
\( \Rightarrow {y_2} = \frac{b}{a}\sqrt {{a^2} – {x^2}} \)
Substituting these values in \((iii),\) we have,
\(A = \int_0^a {\left\{ {\frac{b}{a}\sqrt {{a^2} – {x^2}} – \frac{b}{a}(a – x)} \right\}} dx\)
\( \Rightarrow A = \frac{b}{a}\left[ {\int_0^a {\sqrt {{a^2} – {x^2}} } dx – \int_0^a {(a – x)} dx} \right]\)
\( \Rightarrow A = \frac{b}{a}\left[ {\left[ {\frac{1}{2}x\sqrt {{a^2} – {x^2}} + \frac{1}{2}{a^2} \sin^{-1} \left( {\frac{x}{a}} \right)} \right]_0^a – \left[ {ax – \frac{{{x^2}}}{2}} \right]_0^a} \right]\)
\( \Rightarrow A = \frac{b}{a}\left[ {\left\{ {0 + \frac{1}{2}{a^2} \sin^{-1} (1)} \right\} – \left\{ {{a^2} – \frac{{{a^2}}}{2}} \right\}} \right]\)
\( = \frac{b}{a}\left\{ {\frac{1}{2}{a^2} \times \frac{\pi }{2} – \frac{{{a^2}}}{2}} \right\}\)
\(\therefore A = \frac{1}{2}\left( {\frac{\pi }{2} – 1} \right)\,{\rm{absq}}{\rm{.units}}\)

Q.4. Find the area of the region bounded by the circle \({x^2} + {y^2} = 16\) and the parabola \({x^2} = 6y.\)
Ans: Given equation of the curves are
\({x^2} + {y^2} = 16 \ldots .(i)\)
\({x^2} = 6y…..\left( {ii} \right)\)
\({x^2} + {y^2} = 16\) represents a circle having centre at the origin and radius four units and \({x^2} = 6y\) represents a parabola opening with axis along \(y−\)axis upward and having its vertex at the origin.
To find the point of intersection of these curves,
Put \(y = \frac{{{x^2}}}{6},\) obtained from \((ii)\) in \((i),\) we get
\({x^2} + \frac{{{x^4}}}{{36}} = 16\)
\( \Rightarrow {x^4} + 36{x^2} – 576 = 0\)
\( \Rightarrow \left( {{x^2} + 48} \right)\left( {{x^2} – 12} \right) = 0\)
\( \Rightarrow {x^2} – 12 = 0\)
\( \Rightarrow x = \pm 2\sqrt 3 \)
\(\therefore y = 2\)
Thus, the two curves intersect at \(A(2\sqrt 3 ,2)\) and \(C( – 2\sqrt 3 ,2).\)

Using the horizontal strip method, we know that,
Required Area \({\rm{ = 2[}}\,{\rm{Area}}\,OADO + {\rm{Area}}\,DABD]\)
\( = 2\left[ {\int_0^2 {\left| {{x_1}} \right|} dy + \int_2^4 {\left| {{x_2}} \right|} dy} \right]\)
\( = 2\left[ {\int_0^2 {\sqrt {6y} } dy + \int_2^4 {\sqrt {16 – {y^2}} } dy} \right]\)
\( = 2\left[ {\frac{2}{3}\sqrt 6 \left[ {{y^{\frac{3}{2}}}} \right]_0^2 + \left[ {\frac{1}{2}y\sqrt {16 – {y^2}} + \frac{{16}}{2} \sin^{-1} \frac{y}{4}} \right]_2^4} \right]\)
\( = 2\left[ {\frac{{2\sqrt 6 }}{3} \times (2\sqrt 2 – 0) + \left\{ {8\,\sin^{-1} 1 – \sqrt {16 – 4} – 8\,\sin^{-1}\frac{1}{2}} \right\}} \right]\)
\( = 2\left\{ {\frac{4}{3}\sqrt {12} + 8 \times \frac{\pi }{2} – \sqrt {12} – 8 \times \frac{\pi }{6}} \right\}\)
\( = 2\left( {\frac{8}{3}\sqrt 3 + 4\pi – 2\sqrt 3 – \frac{{4\pi }}{3}} \right)\)
\(\therefore A = \left( {\frac{{4\sqrt 3 }}{3} + \frac{{16\pi }}{3}} \right)\,{\rm{sq}}{\rm{.units}}\)

Q.5. Find the area of the region bounded by the curve \(y = {x^3}\) and the lines \(y=x+6\) and \(y=0.\)
Ans: The region bounded by the curve \(y = {x^3}\) and the lines \(y=x+6\) and \(y=0\) is the shaded region as shown in the figure below.

The approximating rectangle as shown in the figure, has length\( = \left( {{x_2} – {x_1}} \right),\) width\(=dy,\) and area\( = \left( {{x_2} – {x_1}} \right)dy.\)
It can move vertically between \(y=0\) and \(y=8.\) So, the required area \(A\) is given by,
\(A = \int_0^8 {\left( {{x_2} – {x_1}} \right)} dy\)
\(\therefore P\left( {{x_1},y} \right)\) and \(Q\left( {{x_2},y} \right)\) lie on \(y=x+6\) and \(y = {x^3}\) respectively.
\(\therefore y = {x_1} + 6,y = x_2^3\)
Thus, \(A = \int_0^8 {\left\{ {{y^{\frac{1}{3}}} – (y – 6)} \right\}} dy\)
\( = \left[ {\frac{3}{4}{y^{\frac{4}{3}}} – \frac{{{y^2}}}{2} + 6y} \right]_0^8\)
\( = \{ (12 – 32 + 48) – 0\} \)
\(∴A=28\) square units

Summary

Integrating to get the area between two curves is a common use of integration. There are two methods to find the area between two curves, vertical strip method and horizontal strip method. Use the formula, \(A = \int_a^b | f(x) – g(x)|dx\) to find the area bounded by the region between \(y = f(x),y = g(x)\) and on the left and right by the vertical lines \(x=a\) and \(x=b\) respectively, and use the formula, \(A = \int_c^d | \psi (y) – \phi (y)|dy\) to find the area bounded by the region between \(x = \phi (y),x = \psi (y)\) and on the bottom and top by the horizontal lines \(y=c\) and \(y=d\) respectively.

Frequently Asked Questions (FAQs)

Q.1. How do you find the area between two curves?
Ans: For two curves \(y=f(x)\) and \(y=g(x)\) such that \(f(x)>g(x),\) the area between the vertical lines \(x=a\) and \(x=b\) is calculated by the following formula,
\(A = \int_a^b | f(x) – g(x)|dx\)
To find the area of the region between \(x = \phi (y),x = \psi (y)\) and the horizontal lines \(y=c\) and \(y=d,\) use the formula,
\(A = \int_c^d | \psi (y) – \phi (y)|dy\)

Q.2. What does the area between two curves represent?
Ans: Distance is the integral of speed with respect to time. As a result, the area between the two curves reflects how far one particle travelled in comparison to the other.

Q.3. What is the first step toward finding the area between two curves?
Ans: To find the area between two curves, we need to sketch the (rough) graph of the two given curves.

Q.4. Can the area between two curves be negative?
Ans: The formula to find the area between two curves is given as follows:
Let \(y=f(x)\) and \(y=g(x)\) be the two given curves on the interval \([a,b]\) such that \(f(x) \ge g(x)\) then the area is evaluated using the following formula,
\(A = \mathop \smallint \nolimits^b [f(x) – g(x)]dx\)
Similarly, the area between \(x=f(y)\) and \(x=g(y)\) on the interval \([c,d]\) is given by the following formula,
\(A = \int\limits_c^d {[f(y) – g(y)]dy} \)
Thus, the area is an absolute value and will always be positive.

Q.5. How do you find the area between two curves horizontally?
Ans: First, split the supplied region into horizontal strips between the specified boundaries.
Then, sum the areas of the horizontal strips to determine the area of the section between two curves using integration.
The area between two curves \(x=f(y)\) and \(x=g(y)\) such that \(f(y) > g(y)\) bounded by the horizontal lines \(y=c,\) and \(y=d\) is given by the following formula,
Area \(= \int\limits_c^d {[f(y) – g(y)]dy} \).