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 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. 

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