Area Between CurvesMathematics 11: Lecture 41

Dan Sloughter

Furman University

December 3, 2007

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 1 / 15

Example: area between curves

I Let A be the area of the region R bounded by the curves y = x2 andy = x + 2:

−2 −1 1 2 3

1

2

3

4

5

R

I Now x2 = x + 2 when 0 = x2 − x − 2 = (x − 2)(x + 1), that is, whenx = −1 or x = 2.

I Hence the two curves intersect at the points (−1, 1) and (2, 4).

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 2 / 15

Example: area between curves

I Let A be the area of the region R bounded by the curves y = x2 andy = x + 2:

−2 −1 1 2 3

1

2

3

4

5

R

I Now x2 = x + 2 when 0 = x2 − x − 2 = (x − 2)(x + 1), that is, whenx = −1 or x = 2.

I Hence the two curves intersect at the points (−1, 1) and (2, 4).

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 2 / 15

Example: area between curves

I Let A be the area of the region R bounded by the curves y = x2 andy = x + 2:

−2 −1 1 2 3

1

2

3

4

5

R

I Now x2 = x + 2 when 0 = x2 − x − 2 = (x − 2)(x + 1), that is, whenx = −1 or x = 2.

I Hence the two curves intersect at the points (−1, 1) and (2, 4).

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 2 / 15

Example (cont’d)

I Now for −1 ≤ x ≤ 2, R is the region bounded above by the curvey = x + 2 and below by the curve y = x2.

I Since

A1 =

∫ 2−1

(x + 2)dx

is the area beneath y = x + 2 and above [−1, 2] and

A2 =

∫ 2−1

x2dx

is the area beneath y = x2 and above [−1, 2], A = A1 − A2.I That is,

A =

∫ 2−1

(x + 2)dx −∫ 2−1

x2dx =

∫ 2−1

(2 + x − x2)dx .

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 3 / 15

Example (cont’d)

I Now for −1 ≤ x ≤ 2, R is the region bounded above by the curvey = x + 2 and below by the curve y = x2.

I Since

A1 =

∫ 2−1

(x + 2)dx

is the area beneath y = x + 2 and above [−1, 2] and

A2 =

∫ 2−1

x2dx

is the area beneath y = x2 and above [−1, 2], A = A1 − A2.

I That is,

A =

∫ 2−1

(x + 2)dx −∫ 2−1

x2dx =

∫ 2−1

(2 + x − x2)dx .

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 3 / 15

Example (cont’d)

I Now for −1 ≤ x ≤ 2, R is the region bounded above by the curvey = x + 2 and below by the curve y = x2.

I Since

A1 =

∫ 2−1

(x + 2)dx

is the area beneath y = x + 2 and above [−1, 2] and

A2 =

∫ 2−1

x2dx

is the area beneath y = x2 and above [−1, 2], A = A1 − A2.I That is,

A =

∫ 2−1

(x + 2)dx −∫ 2−1

x2dx =

∫ 2−1

(2 + x − x2)dx .

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 3 / 15

Example (cont’d)

I Hence

A = 2x

∣∣∣∣2−1

+1

2x2

∣∣∣∣∣2

−1

− 13x3

∣∣∣∣∣2

−1

= (4 + 2) +

(2− 1

2

)−

(8

3+

1

3

)= 5− 1

2

=9

2

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 4 / 15

Area between curves

I In general, if f (x) ≤ g(x) for all x in the interval [a, b] and A is thearea of the region R which lies between the curves y = f (x) andy = g(x) over the interval [a, b], then

A =

∫ ba

(g(x)− f (x))dx .

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 5 / 15

ExampleI Let A be the area of the region R bounded by the curves y = x3 − x2

and y = 2x :

−2 −1 1 2 3

−2

2

4

R2

R1

I First we see that the two curves intersect when x3 − x2 = 2x , that is,when

0 = x3 − x2 − 2x = x(x2 − x − 2) = x(x − 2)(x + 1).

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 6 / 15

ExampleI Let A be the area of the region R bounded by the curves y = x3 − x2

and y = 2x :

−2 −1 1 2 3

−2

2

4

R2

R1

I First we see that the two curves intersect when x3 − x2 = 2x , that is,when

0 = x3 − x2 − 2x = x(x2 − x − 2) = x(x − 2)(x + 1).

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 6 / 15

Example (cont’d)

I Hence the curves intersect at (−1,−2), (0, 0) and (2, 4).

I Now for −1 < x < 0, x3 − x2 > 2x , and for 0 < x < 2, x3 − x2 < 2x .I Break R into two regions:

I R1, bounded above by y = x3 − x2 and below by y = 2x for

−1 ≤ x ≤ 0,I R2, bounded above by y = 2x and below by y = x

3 − x2 for 0 ≤ x ≤ 2.

I Hence

A =

∫ 0−1

(x3 − x2 − 2x)dx +∫ 2

0(2x − x3 + x2)dx

=1

4x4

∣∣∣∣∣0

−1

− 13x3

∣∣∣∣∣0

−1

− x2∣∣∣∣0−1

+ x2∣∣∣∣20

− 14x4

∣∣∣∣∣2

0

+1

3x3

∣∣∣∣∣2

0

= −14− 1

3+ 1 + 4− 4 + 8

3=

37

12.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 7 / 15

Example (cont’d)

I Hence the curves intersect at (−1,−2), (0, 0) and (2, 4).I Now for −1 < x < 0, x3 − x2 > 2x , and for 0 < x < 2, x3 − x2 < 2x .

I Break R into two regions:

I R1, bounded above by y = x3 − x2 and below by y = 2x for

−1 ≤ x ≤ 0,I R2, bounded above by y = 2x and below by y = x

3 − x2 for 0 ≤ x ≤ 2.

I Hence

A =

∫ 0−1

(x3 − x2 − 2x)dx +∫ 2

0(2x − x3 + x2)dx

=1

4x4

∣∣∣∣∣0

−1

− 13x3

∣∣∣∣∣0

−1

− x2∣∣∣∣0−1

+ x2∣∣∣∣20

− 14x4

∣∣∣∣∣2

0

+1

3x3

∣∣∣∣∣2

0

= −14− 1

3+ 1 + 4− 4 + 8

3=

37

12.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 7 / 15

Example (cont’d)

I Hence the curves intersect at (−1,−2), (0, 0) and (2, 4).I Now for −1 < x < 0, x3 − x2 > 2x , and for 0 < x < 2, x3 − x2 < 2x .I Break R into two regions:

I R1, bounded above by y = x3 − x2 and below by y = 2x for

−1 ≤ x ≤ 0,I R2, bounded above by y = 2x and below by y = x

3 − x2 for 0 ≤ x ≤ 2.

I Hence

A =

∫ 0−1

(x3 − x2 − 2x)dx +∫ 2

0(2x − x3 + x2)dx

=1

4x4

∣∣∣∣∣0

−1

− 13x3

∣∣∣∣∣0

−1

− x2∣∣∣∣0−1

+ x2∣∣∣∣20

− 14x4

∣∣∣∣∣2

0

+1

3x3

∣∣∣∣∣2

0

= −14− 1

3+ 1 + 4− 4 + 8

3=

37

12.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 7 / 15

Example (cont’d)

I Hence the curves intersect at (−1,−2), (0, 0) and (2, 4).I Now for −1 < x < 0, x3 − x2 > 2x , and for 0 < x < 2, x3 − x2 < 2x .I Break R into two regions:

I R1, bounded above by y = x3 − x2 and below by y = 2x for

−1 ≤ x ≤ 0,

I R2, bounded above by y = 2x and below by y = x3 − x2 for 0 ≤ x ≤ 2.

I Hence

A =

∫ 0−1

(x3 − x2 − 2x)dx +∫ 2

0(2x − x3 + x2)dx

=1

4x4

∣∣∣∣∣0

−1

− 13x3

∣∣∣∣∣0

−1

− x2∣∣∣∣0−1

+ x2∣∣∣∣20

− 14x4

∣∣∣∣∣2

0

+1

3x3

∣∣∣∣∣2

0

= −14− 1

3+ 1 + 4− 4 + 8

3=

37

12.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 7 / 15

Example (cont’d)

I Hence the curves intersect at (−1,−2), (0, 0) and (2, 4).I Now for −1 < x < 0, x3 − x2 > 2x , and for 0 < x < 2, x3 − x2 < 2x .I Break R into two regions:

I R1, bounded above by y = x3 − x2 and below by y = 2x for

−1 ≤ x ≤ 0,I R2, bounded above by y = 2x and below by y = x

3 − x2 for 0 ≤ x ≤ 2.

I Hence

A =

∫ 0−1

(x3 − x2 − 2x)dx +∫ 2

0(2x − x3 + x2)dx

=1

4x4

∣∣∣∣∣0

−1

− 13x3

∣∣∣∣∣0

−1

− x2∣∣∣∣0−1

+ x2∣∣∣∣20

− 14x4

∣∣∣∣∣2

0

+1

3x3

∣∣∣∣∣2

0

= −14− 1

3+ 1 + 4− 4 + 8

3=

37

12.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 7 / 15

Example (cont’d)

I Hence the curves intersect at (−1,−2), (0, 0) and (2, 4).I Now for −1 < x < 0, x3 − x2 > 2x , and for 0 < x < 2, x3 − x2 < 2x .I Break R into two regions:

I R1, bounded above by y = x3 − x2 and below by y = 2x for

−1 ≤ x ≤ 0,I R2, bounded above by y = 2x and below by y = x

3 − x2 for 0 ≤ x ≤ 2.

I Hence

A =

∫ 0−1

(x3 − x2 − 2x)dx +∫ 2

0(2x − x3 + x2)dx

=1

4x4

∣∣∣∣∣0

−1

− 13x3

∣∣∣∣∣0

−1

− x2∣∣∣∣0−1

+ x2∣∣∣∣20

− 14x4

∣∣∣∣∣2

0

+1

3x3

∣∣∣∣∣2

0

= −14− 1

3+ 1 + 4− 4 + 8

3=

37

12.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 7 / 15

Example

I We will find the area A of the region R bounded by the curves x = y2

and y = 2− x .

I Note: these curves intersect when y2 = 2− y , that is when

0 = y2 + y − 2 = (y + 2)(y − 1).

I Thus the curves intersect at the points (1, 1) and (4,−2).I Note: over the interval [0, 1], R is bounded above by y =

√x and

below by y = −√

x and, over the interval [1, 4], R is bounded aboveby y = 2− x and below by y = −

√x .

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 8 / 15

Example

I We will find the area A of the region R bounded by the curves x = y2

and y = 2− x .I Note: these curves intersect when y2 = 2− y , that is when

0 = y2 + y − 2 = (y + 2)(y − 1).

I Thus the curves intersect at the points (1, 1) and (4,−2).I Note: over the interval [0, 1], R is bounded above by y =

√x and

below by y = −√

x and, over the interval [1, 4], R is bounded aboveby y = 2− x and below by y = −

√x .

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 8 / 15

Example

I We will find the area A of the region R bounded by the curves x = y2

and y = 2− x .I Note: these curves intersect when y2 = 2− y , that is when

0 = y2 + y − 2 = (y + 2)(y − 1).

I Thus the curves intersect at the points (1, 1) and (4,−2).

I Note: over the interval [0, 1], R is bounded above by y =√

x andbelow by y = −

√x and, over the interval [1, 4], R is bounded above

by y = 2− x and below by y = −√

x .

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 8 / 15

Example

I We will find the area A of the region R bounded by the curves x = y2

and y = 2− x .I Note: these curves intersect when y2 = 2− y , that is when

0 = y2 + y − 2 = (y + 2)(y − 1).

I Thus the curves intersect at the points (1, 1) and (4,−2).I Note: over the interval [0, 1], R is bounded above by y =

√x and

below by y = −√

x and, over the interval [1, 4], R is bounded aboveby y = 2− x and below by y = −

√x .

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 8 / 15

Example (cont’d)

I Hence we have

A =

∫ 10

(√

x − (−√

x))dx +

∫ 41

(2− x − (−√

x))dx

= 2

∫ 10

√xdx +

∫ 41

(2− x +√

x)dx

=4

3x

32

∣∣∣∣∣1

0

+ 2x

∣∣∣∣41

− 12x2

∣∣∣∣∣4

1

+2

3x

32

∣∣∣∣∣4

1

=4

3+ (8− 2)−

(8− 1

2

)+

(16

3− 2

3

)=

9

2.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 9 / 15

Example (cont’d)

I The region between the curves x = y2 and y = 2− x :

1 2 3 4 5

−3

−2

−1

1

2

R1

R2

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 10 / 15

Another approach

I Note: if A is the area of the region bounded by the curves x = g(y)and x = f (y) over an interval [c , d ], where we assume f (y) ≤ g(y)for all y in [c , d ], then, analogous to our previous formula,

A =

∫ dc

(g(y)− f (y))dy .

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 11 / 15

Example (cont’d)

I We may also find the area A of the region R by

A =

∫ 1−2

(2− y − y2)dy

= 2y

∣∣∣∣1−2− 1

2y2

∣∣∣∣∣1

−2

− 13y3

∣∣∣∣∣1

−2

= (2 + 4)−(

1

2− 2

)−

(1

3+

8

3

)=

9

2.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 12 / 15

Example

I Let A be the area of the region R bounded by x = y2 and x = y .

I Now y = y2 when y(y − 1) = 0, that is, when y = 0, or y = 1.I Hence the curves intersect at (0, 0), and (1, 1).

I Thus

A =

∫ 10

(y − y2)dy = 12y2

∣∣∣∣∣1

0

− 13y3

∣∣∣∣∣1

0

=1

2− 1

3=

1

6.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 13 / 15

Example

I Let A be the area of the region R bounded by x = y2 and x = y .

I Now y = y2 when y(y − 1) = 0, that is, when y = 0, or y = 1.

I Hence the curves intersect at (0, 0), and (1, 1).

I Thus

A =

∫ 10

(y − y2)dy = 12y2

∣∣∣∣∣1

0

− 13y3

∣∣∣∣∣1

0

=1

2− 1

3=

1

6.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 13 / 15

Example

I Let A be the area of the region R bounded by x = y2 and x = y .

I Now y = y2 when y(y − 1) = 0, that is, when y = 0, or y = 1.I Hence the curves intersect at (0, 0), and (1, 1).

I Thus

A =

∫ 10

(y − y2)dy = 12y2

∣∣∣∣∣1

0

− 13y3

∣∣∣∣∣1

0

=1

2− 1

3=

1

6.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 13 / 15

Example

I Let A be the area of the region R bounded by x = y2 and x = y .

I Now y = y2 when y(y − 1) = 0, that is, when y = 0, or y = 1.I Hence the curves intersect at (0, 0), and (1, 1).

I Thus

A =

∫ 10

(y − y2)dy = 12y2

∣∣∣∣∣1

0

− 13y3

∣∣∣∣∣1

0

=1

2− 1

3=

1

6.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 13 / 15

Example (cont’d)

I Note: we could also find A as follows:

A =

∫ 10

(√

x − x)dy = 23x

32

∣∣∣∣∣1

0

− 12x2

∣∣∣∣∣1

0

=2

3− 1

2=

1

6.

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 14 / 15

Example (cont’d)

I Region R bounded by the curves x = y2 and x = y :

−0.2 0.2 0.4 0.6 0.8 1 1.2

−1

−0.5

0.5

1

R

Dan Sloughter (Furman University) Area Between Curves December 3, 2007 15 / 15