Problems on Area Between Two Curves
Problem 1: Find the area bounded between two lines f(x) = 5x and g(x) = 3x from x =0 to x = 3.
Solution:
The figure below shows both the lines,
From the figure, we know
Area = [Tex]\int^{b}_a[f(x) – g(x)]dx [/Tex]
⇒ Area = [Tex]\int^{3}_0[f(x) – g(x)]dx [/Tex]
⇒ Area = [Tex]\int^{3}_0[5x – 3x]dx [/Tex]
⇒ Area = [Tex]\int^{3}_0[2x]dx [/Tex]
⇒ Area = [Tex][x^2]^3_0 [/Tex]
⇒ Area = 9 sq. units
Problem 2: Find the area bounded between two curves f(x) = x3 and g(x) = x2 between 0 and 1.
Solution:
The figure below shows both the curves, to find the bounded region, we first need to find the intersections.
f(x) = g(x)
⇒x3 = x2
⇒x2(x-1) = 0
⇒ x = 0 and 1
From the figure, we know
Area = [Tex]\int^{b}_a[f(x) – g(x)]dx [/Tex]
⇒ Area = [Tex]\int^{1}_0[f(x) – g(x)]dx [/Tex]
⇒ Area = [Tex]\int^{3}_0[x^2 – x^3]dx [/Tex]
⇒ Area = [Tex]\int^{1}_{0}x^2dx – \int^1_0x^3dx [/Tex]
⇒ Area = [Tex][\frac{x^3}{3}]^1_0 – [\frac{x^4}{4}]^1_0 [/Tex]
⇒ Area = 1/3 – 1/4
⇒ Area = 1/12 sq. units
Problem 3: Find the area bounded between the parabola y2 = 4x and x2 + y2 = 9.
Solution:
The figure below shows both the curves, to find the bounded region, we first need to find the intersections.
x2 + y2 = 12
Figure
Area = [Tex]\int^{b}_a[f(x) – g(x)]dx [/Tex]
⇒ Area = [Tex]\int^{1}_0[f(x) – g(x)]dx [/Tex]
⇒ Area = [Tex]\int^{1}_0[x – x^2]dx [/Tex]
⇒ Area = [Tex]\int^{1}_{0}xdx – \int^1_0x^2dx [/Tex]
⇒ Area = [Tex][\frac{x^2}{2}]^1_0 – [\frac{x^3}{3}]^1_0 [/Tex]
⇒ Area = 1/2 – 1/3
⇒ Area = 1/6 sq. units
Problem 4: Find the area bounded between the parabola y2 = 4x and its latus rectum.
Solution:
The figure below shows the parabola, and it’s latus rectum. Latus rectum is the line x = 1. We need to find the intersections,
y2 = 4
y = 2 and -2
Area = 2(Area of the region bounded by the parabola and x = 1 and x-axis in the first quadrant)
⇒ Area = 2([Tex]\int^1_0ydx [/Tex])
⇒ Area = 2[Tex]\int^1_0\sqrt{4x}dx [/Tex]
⇒ Area = 4[Tex]\int^1_0\sqrt{x}dx [/Tex]
⇒ Area = 4[Tex]\int^1_0[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]dx [/Tex]
⇒ Area = [Tex]\frac{8}{3}[x^{\frac{3}{2}}]^1_0 [/Tex]
⇒ Area = 8/3 sq. units
Problem 5: The figure given below shows an ellipse 9x2+ y2 = 36 and a chord PQ. Find the area enclosed between the chord and the ellipse in the first quadrant.
Solution:
The equation of ellipse is,
[Tex]\frac{x^2}{4} + \frac{y^2}{36} = 1 [/Tex]
⇒ [Tex]\frac{x^2}{2^2} + \frac{y^2}{6^2} = 1 [/Tex]
So, now the equation of the chord becomes,
⇒ [Tex]\frac{x}{2} + \frac{y}{6} = 1 [/Tex]
⇒ 3x + y = 6
⇒ y = 6 – 3x
So, now the required area will be.
A = [Tex]3\int^{2}_0 \sqrt{4 – x^2}dx – \int^{2}_{0}(6 – 3x)dx [/Tex]
⇒ A= [Tex]3[\frac{x}{2}\sqrt{4 – x^2} + 2sin^{-1}\frac{x}{2}]^2_0 – [6x – \frac{3x^2}{2}]^2_0 [/Tex]
⇒ A= [Tex]6sin^{-1}(1) – 6 [/Tex]
⇒ A = 3π – 6 sq. units
Area Between Two Curves: Formula, Definition and Examples
Area Between Two Curves in Calculus is one of the applications of Integration. It helps us calculate the area bounded between two or more curves using the integration. As we know Integration in calculus is defined as the continuous summation of very small units. The topic “Area Between Two Curves” has applications in the various fields of engineering, physics, and economics.
Let’s know more about Area Between Two Curves in detail below.