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6.2 Regions Between Curves

The area under \( y = f(x) \) above the \( x \)-axis between \( x = a \) and \( x = b \) is

\[ \int_{a}^{b} f(x) \, dx \]

area of a thin rectangle

height \( f(x) \)

width \( dx \)

Graph of a curve y=f(x) above the x-axis from a to b, showing a thin vertical rectangle of width dx and height f(x).

Sum of infinitely many \( f(x) \, dx \) as \( x \) goes from \( a \) to \( b \).

This idea can be extended to find area between two curves

Graph showing two curves y=f(x) and y=g(x) with a vertical rectangle between them of height f(x)-g(x) and width dx.

each rectangle has area = \( [f(x) - g(x)] \, dx \)

use integration to accumulate

\[ \int_{a}^{b} [\underbrace{f(x)}_{\text{top}} - \underbrace{g(x)}_{\text{bottom}}] \, dx \]
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Example

Find the area of region bounded by

\( y = x^2 - 2x \)

parabola opens up

\( y = 2x \)

line

Graph of a parabola y=x^2-2x and a line y=2x intersecting at (0,0) and (4,8), with the area between them shaded.

\( y = x^2 - 2x \)

\( = x(x - 2) \rightarrow x \)-intercepts \( (y=0) \)

at \( x = 0, x = 2 \)

the left end of the region:

the right end of the region:

intersections of \( y = x^2 - 2x \) and \( y = 2x \)

\[ x^2 - 2x = 2x \]\[ x^2 - 4x = 0 \rightarrow x(x - 4) = 0 \]\[ x = 0, x = 4 \]

rectangle height: \( 2x - (x^2 - 2x) = 4x - x^2 \)

rectangle width: \( dx \)

Area of region:

\[ \int_{0}^{4} (4x - x^2) \, dx = 4 \cdot \frac{x^2}{2} - \frac{x^3}{3} \bigg|_0^4 = 2x^2 - \frac{1}{3}x^3 \bigg|_0^4 = (32 - \frac{64}{3}) - 0 = \frac{32}{3} \]
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Example: Area Between Curves

Find the area of the region bound by \( y = \cos x \) and \( y = \sin 2x \) between \( x = 0 \) and \( x = \frac{\pi}{2} \).

Coordinate graph showing y=cos(x) and y=sin(2x) intersecting at x=pi/6 and x=pi/2.
  • Region â‘ : \( \cos x \) is above, \( \sin 2x \) is below.
  • Region â‘¡: \( \sin 2x \) is above, \( \cos x \) is below.

Calculations for bounds:

\( \sin \frac{\pi}{2} = 1 \)

\( \sin 2x = 1 \rightarrow 2x = \frac{\pi}{2} \)

\( x = \frac{\pi}{4} \)

Finding Intersection Points

Intersection:

\[ \cos x = \sin 2x \]\[ \cos x = 2 \sin x \cos x \]\[ \cos x - 2 \sin x \cos x = 0 \]\[ \cos x (1 - 2 \sin x) = 0 \]\[ \cos x = 0 \implies x = \frac{\pi}{2} \]\[ 1 - 2 \sin x = 0 \implies \sin x = \frac{1}{2} \implies x = \frac{\pi}{6} \]

Identity:

\[ \sin 2x = 2 \sin x \cos x \]
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Definite Integral Setup

\[ \underbrace{\int_{0}^{\pi/6} (\cos x - \sin 2x) \, dx}_{\text{Region â‘ }} + \underbrace{\int_{\pi/6}^{\pi/2} (\sin 2x - \cos x) \, dx}_{\text{Region â‘¡}} \]

Recall integration rules:

\[ \int \cos x \, dx = \sin x + C \]\[ \int \sin x \, dx = -\cos x + C \]

Evaluation

\[ = \left. \left( \sin x + \frac{1}{2} \cos 2x \right) \right|_{0}^{\pi/6} + \left. \left( -\frac{1}{2} \cos 2x - \sin x \right) \right|_{\pi/6}^{\pi/2} \]\[ = \left( \sin \frac{\pi}{6} + \frac{1}{2} \cos \frac{\pi}{3} \right) - \left( \sin 0 + \frac{1}{2} \cos 0 \right) + \left( -\frac{1}{2} \cos \pi - \sin \frac{\pi}{2} \right) - \left( -\frac{1}{2} \cos \frac{\pi}{3} - \sin \frac{\pi}{6} \right) \]\[ = \left( \frac{1}{2} + \frac{1}{4} \right) - \left( 0 + \frac{1}{2} \right) + \left( \frac{1}{2} - 1 \right) - \left( -\frac{1}{4} - \frac{1}{2} \right) \]
\[ = \frac{1}{2} \]
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Integration in Terms of y

If we can integrate in terms of \(x\), we can integrate in terms of \(y\).

Coordinate graph showing area between curves x=g(y) and x=f(y) from y=c to y=d using horizontal rectangles.

Figure 1: Area between curves with respect to y.

Width: \(f(y) - g(y)\)
right - left

\[ \text{area} = \int_{c}^{d} [f(y) - g(y)] \, dy \]

Example

\(x = y^2 - 6\), \(x = y\)

  • \(x = y^2 - 6\) is a parabola opening right.
  • \(x = y\) is a line.
Graph of x=y^2-6 and x=y showing vertical rectangles that change top/bottom boundaries at the vertex.

Figure 2: Intersection of parabola and line with vertical rectangles.

\(x = y^2 - 6\) find y-ints (\(x=0\))

\(0 = y^2 - 6 \implies y = \pm \sqrt{6}\)

If we use vertical rectangles, then the roles of "top" and "bottom" switch/change at some point.

→ ok, but maybe there is something easier

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Horizontal rectangles are a little easier

Graph of parabola x=y^2-6 and line x=y with horizontal rectangles from y=-2 to y=3.

Figure 3: Area calculation using horizontal rectangles.

This time, "right" is always \(x = y\)

"left" is always \(x = y^2 - 6\)

no switching!

Intersections:

\[ y^2 - 6 = y \]

\[ y^2 - y - 6 = 0 \]

\[ (y - 3)(y + 2) = 0 \]

\[ y = -2, \, y = 3 \]

\[ \int_{-2}^{3} [\underbrace{y}_{\text{right}} - (\underbrace{y^2 - 6}_{\text{left}})] \, dy = \int_{-2}^{3} (y - y^2 + 6) \, dy \]

\[ = \frac{y^2}{2} - \frac{y^3}{3} + 6y \Big|_{-2}^{3} = \dots = \boxed{\frac{125}{6}} \]