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3. Area Between 2 Curves

by M. Bourne

We are trying to find the area between 2 curves, y₁ = f₁(x) and y₂ = f₂(x) and the lines x = a and x = b.

We see that if we subtract the area under lower curve

y₁ = f₁(x)

from the area under the upper curve

y₂ = f₂(x),

then we will find the required area. This can be achieved in one step:

Alternative Way to Find The Formula (First Principles)

Another way of deriving this formula is as follows (the thinking here is important for understanding how we develop the later formulas in this section).

Each "typical" rectangle indicated has width Δx and height y₂ − y₁, so its area is (y₂ − y₁)Δx.

If we add all these typical rectangles, starting from a and finishing at b, the area is approximately:

Now if we let Δx → 0, we can find the exact area by integration:

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Likewise, we can sum vertically by re-expressing both functions so that they are functions of y and we find:

 

Example:

Find the area between the curves y = x² + 5x and y = 3 − x² between x = -2 and x = 0.

Answer

Sketching first:

So we need to find:

 

Exercises

1. Find the area bounded by y = x³, x = 0 and y = 3.

 

Answer

Sketch first:

We need to use:

and use horizontal elements.

In this case, c = 0 and d = 3.

We need to express x in terms of y:

y = x³ so x = y^1/3

So

 

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2. Find the area bounded by the curves

y = x² + 5x and y = 3 − x².

(This is an extension of the Example above.)

Here is the answer using LiveMath:

LIVEMath

Normal answer:

Answer

Sketch first:

We need to use:

We note that y = 3 − x² is above y = x²+ 5x so we take

y₂ = 3 − x² and y₁= x² + 5x

Points of intersection occur where:

x² + 5x = 3 − x²

2x² + 5x − 3 = 0

(x + 3)(2x − 1) = 0

So x = -3 or x = 0.5

We take vertical elements.

So the area is given by:

 

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3. Find the area bounded by the curves

y = x², y = 2 − x and y = 1.

Answer

Sketch first:

We need to take horizontal elements in this case.

So we need to solve y = x² for x:

x = ±√y

We need the left hand portion, so x = − √y.

Notice that x = 2 − y is to the right of x = -√y so we choose
x₂ = 2 − y and x₁ = -√y.

The intersection of the graphs occurs at (-2,4) and (1,1).

So we have: c = 1 and d = 4.

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Chapter Contents

• Applications of Integration
• 1. Applications of the Indefinite Integral
• 2. Area Under a Curve
• 3. Area Between 2 Curves
• 4. Volume of Solid of Integration
• 5. Centroid of an Area
• 6. Moments of Inertia
• 7. Work by a Variable Force
• 8. Electric Charges
• 9. Average Value
• Head Injury Criterion
• 10. Force by Liquid Pressure

• Comments, Questions?

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