MATH 12002 - CALCULUS I 7.1: Area Between Curves Professor Donald - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §7.1: Area Between Curves

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Area Between Curves

We saw previously that the area of the region bounded by the graph of a function and the x-axis can be computed in term of definite integrals. We now generalize this to compute the area of a region bounded by the graphs of two functions, such as the area of the shaded region S in the figure below:

§7.1 FIGURE 1

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Area Between Curves

Theorem

If f (x) and g(x) are continuous functions such that f (x) g(x)

• n the interval [a, b], then the area of the region bounded by the graphs of

f and g and the lines x = a and x = b is A = b

a

[f (x) − g(x)] dx. The formula applies only when f (x) g(x) for all x in [a, b]. We will see later how to adapt it to more general situations. If f and g are both positive on [a, b], then the area A between the graphs of f and g is the area under the graph of f , that is, b

a f (x) dx,

minus the area under the graph of g, that is b

a g(x) dx, hence

A = b

a

f (x) dx − b

a

g(x) dx = b

a

[f (x) − g(x)] dx. The formula is proved in general using Riemann sums in the text.

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Examples

1 Find the area of the region between the graphs of y = x2 and y = x

• n the interval [0, 1].

We need to find the area of the shaded region in the figure below: Since x x2 for 0 x 1, the area is A = 1 x − x2 dx = 1 2x2 − 1 3x3

• 1

= 1 2 − 1 3 = 1 6.

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Examples

If two graphs intersect at two or more points, we can find the area of the region bounded by the two graphs between their first (left-most) intersection point and their last (right-most) intersection point.

2 Find the area of the region bounded by the graphs of y = x2 − 3 and

y = 5 − x2. We first need to determine where the graphs intersect. An intersection point occurs at an x-value where the y-values are equal, so we equate the two y-values and solve the equation for x: x2 − 3 = 5 − x2, 2x2 − 8 = 0, 0 = 2(x2 − 4) = 2(x − 2)(x + 2). Therefore, the two graphs intersect at x = −2 and x = 2.

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Examples

[Example 2, continued] We therefore want to find the area of the shaded region in the figure below: Since 5 − x2 x2 − 3 for −2 x 2, the area is A = 2

−2

(5 − x2) − (x2 − 3) dx = 2

−2

8 − 2x2 dx.

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Examples

[Example 2, continued] Finally, we have A = 2

−2

8 − 2x2 dx =

• 8x − 2

3x3

• 2

−2

=

• 8(2) − 2

3(23)

• −
• 8(−2) − 2

3((−2)3)

• =
• 16 − 16

3

• −
• −16 + 16

3

• =

16 − 16 3 + 16 − 16 3 = 32 − 32 3 = 64 3 .

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Examples

3 Find the area of the region bounded by the graphs of y = x2 − 2x

and y = x + 4. We first determine the points of intersection. We set x2 − 2x = x + 4, so that 0 = x2 − 3x − 4 = (x + 1)(x − 4), and so the graphs intersect at x = −1 and x = 4. Hence we want to find the area of the shaded region in the following figure:

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Examples

[Example 3, continued] Since x + 4 x2 − 2x for −1 x 4, the area is A = 4

−1

(x + 4) − (x2 − 2x) dx = 4

−1

4 + 3x − x2 dx.

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Examples

[Example 3, continued] Finally, we have A = 4

−1

4 + 3x − x2 dx =

• 4x + 3

2x2 − 1 3x3

• 4

−1

=

• 4(4) + 3

2(42) − 1 3(43)

• −
• 4(−1) + 3

2((−1)2) − 1 3((−1)3)

• =
• 16 + 24 − 64

3

• −
• −4 + 3

2 + 1 3

• =

40 − 64 3 + 4 − 3 2 − 1 3 = 125 6 .

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