Math 220Calculus I

Section 6.4 Areas in the x-y Plane

Example F

6.4 Example A-1

Find the area of the geometric figure pictured below.

2 3

6.4 Example A-1

Find the area of the geometric figure pictured below.

2 3

The figure consists of a rectangle and a semi-circle.

6.4 Example A-1

Find the area of the geometric figure pictured below.

2 3

The figure consists of a rectangle and a semi-circle. (Note that the width of the rectangle, 2, is the same as the radius of the semi-circle.)

6.4 Example A-1

Find the area of the geometric figure pictured below.

2 3

The figure consists of a rectangle and a semi-circle. (Note that the width of the rectangle, 2, is the same as the radius of the semi-circle.)

The total area is the sum of the two areas.

6.4 Example A-1

Find the area of the geometric figure pictured below.

2 3

The figure consists of a rectangle and a semi-circle. (Note that the width of the rectangle, 2, is the same as the radius of the semi-circle.)

The total area is the sum of the two areas.

222

132 A

6.4 Example A-1

Find the area of the geometric figure pictured below.

2 3

The figure consists of a rectangle and a semi-circle. (Note that the width of the rectangle, 2, is the same as the radius of the semi-circle.)

The total area is the sum of the two areas.

€

A = 2 3( ) +1

2π 22

( )

A = 6 + 2π

6.4 Theory

In Lectures 6.2 and 6.3 we were able to equate the area under a curve on an interval a ≤ x ≤ b, with the definite integral from a to b:

Combining this concept with geometric ideas such as those used in Example A supports the assertion of some of the properties of integrals already used in Lecture 6.1.

b

adxxf

6.4 Theory

In Lectures 6.2 and 6.3 we were able to equate the area under a curve on an interval a ≤ x ≤ b, with the definite integral from a to b:

Combining this concept with geometric ideas such as those used in Example A supports the assertion of some of the properties of integrals already used in Lecture 6.1.

b

adxxf

b

a

b

a

b

adxxgdxxfdxxgxf

6.4 Theory

In Lectures 6.2 and 6.3 we were able to equate the area under a curve on an interval a ≤ x ≤ b, with the definite integral from a to b:

Combining this concept with geometric ideas such as those used in Example A supports the assertion of some of the properties of integrals already used in Lecture 6.1.

b

adxxf

b

a

b

a

b

adxxgdxxfdxxgxf

b

a

b

a

b

adxxgdxxfdxxgxf

6.4 Theory

In Lectures 6.2 and 6.3 we were able to equate the area under a curve on an interval a ≤ x ≤ b, with the definite integral from a to b:

Combining this concept with geometric ideas such as those used in Example A supports the assertion of some of the properties of integrals already used in Lecture 6.1.

b

adxxf

b

a

b

a

b

adxxgdxxfdxxgxf

b

a

b

a

b

adxxgdxxfdxxgxf

b

a

b

adxxfkdxxfk

6.4 Example B

Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3.

x

xg1

6.4 Example B

Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3.

x

xg1

As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

6.4 Example B

Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3.

x

xg1

As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

3

1

11 dx

xx

“above”

“below”

6.4 Example B

Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3.

x

xg1

As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

1

323

1ln

2

111 xxxdx

xx

6.4 Example B

Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3.

x

xg1

As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

1ln112

13ln33

2

1

ln2

111

22

1

323

1xxxdx

xx

6.4 Example B

Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3.

x

xg1

As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

012

13ln3

2

9

1ln112

13ln33

2

1

ln2

111

22

1

323

1

xxxdxx

x

6.4 Example B

Find the area between the curves f(x) = x + 1 and on the interval 1 ≤ x ≤ 3.

x

xg1

As long as f(x) > g(x) for all values of x in the interval, we can use the second property above and subtract the areas.

3ln6

012

13ln3

2

9

1ln112

13ln33

2

1

1

3ln

2

111

22

23

1

xxxdxx

x

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

21

xxg

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

Looking at the graph, the shape remains the same because the curves from Example B were simply shifted down by 2. The area should be the same.

21

xxg

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

Looking at the graph, the shape remains the same because the curves from Example B were simply shifted down by 2. The area should be the same. However, in this case some of the area lies below the x-axis, and would thus be a negative area (as noted in Lecture 6.3).

21

xxg

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

Looking at the graph, the shape remains the same because the curves from Example B were simply shifted down by 2. The area should be the same. However, in this case some of the area lies below the x-axis, and would thus be a negative area (as noted in Lecture 6.3).

21

xxg

Does this affect the calculation of area between the two curves?

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

21

xxg

3

12

11 dx

xx

“above”“below”

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

21

xxg

1

323

12ln

2

12

11 xxxxdx

xx

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

21

xxg

1

32

1

323

1

ln2

1

2ln2

12

11

xxx

xxxxdxx

x

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

21

xxg

1ln112

13ln33

2

1

ln2

1

2ln2

12

11

22

1

32

1

323

1

xxx

xxxxdxx

x

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

21

xxg

012

13ln3

2

9

1ln112

13ln33

2

1

ln2

1

2ln2

12

11

22

1

32

1

323

1

xxx

xxxxdxx

x

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

21

xxg

3ln6012

13ln3

2

9

1ln112

13ln33

2

1

ln2

1

2ln2

12

11

22

1

32

1

323

1

xxx

xxxxdxx

x

6.4 Example C

Find the area between the curves f(x) = x – 1 and on the interval 1 ≤ x ≤ 3.

21

xxg

So the integral calculation gives us the same result as in Example B:

Area = 6 – ln3.

It doesn’t matter whether the two curves lie above or below the x-axis, only that we subtract the “higher/above” curve minus the “lower/below”.

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

Looking at the graph, y = x2 – 4x + 12 lies above y = x2 for a portion of the interval, but is below for the rest of the interval.

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

Looking at the graph, y = x2 – 4x + 12 lies above y = x2 for a portion of the interval, but is below for the rest of the interval.

We need to determine where the two intersect, then set up two integrals: one for each portion.

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

We need to determine where the two intersect, then set up two integrals: one for each portion.

12422 xxx

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

We need to determine where the two intersect, then set up two integrals: one for each portion.

124

12422

x

xxx

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

We need to determine where the two intersect, then set up two integrals: one for each portion.

3

124

12422

x

x

xxx

x = 3

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

x = 3

€

x 2 − 4x +12( ) − x 2[ ] dx

0

3

∫

+ x 2 − x 2 − 4x +12( )[ ] dx3

4

∫

“above”

“below”

“above” “below”

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

x = 3

€

x 2 − 4 x +12( ) − x 2[ ] dx

0

3

∫

+ x 2 − x 2 − 4x +12( )[ ] dx3

4

∫

= −4 x +12[ ] dx0

3

∫ + 4x −12[ ] dx3

4

∫

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

x = 3

€

x 2 − 4x +12( ) − x 2[ ] dx

0

3

∫

+ x 2 − x 2 − 4 x +12( )[ ] dx3

4

∫

= −4x +12[ ] dx0

3

∫ + 4x −12[ ] dx3

4

∫

= −2x 2 +12x[ ]3

0+ 2x 2 −12x[ ]

4

3

Combining like terms is not an option—the two polynomials have different boundaries to evaluate.

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

x = 3

€

x 2 − 4 x +12( ) − x 2[ ] dx

0

3

∫

+ x 2 − x 2 − 4 x +12( )[ ] dx3

4

∫

= −4 x +12[ ] dx0

3

∫ + 4x −12[ ] dx3

4

∫

= −2x 2 +12x[ ]3

0+ 2x 2 −12x[ ]

4

3

= −2 32( ) +12 3( )[ ] − −2 02

( ) +12 0( )[ ] + 2 42( ) −12 4( )[ ] − 2 32

( ) −12 3( )[ ]

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

x = 3

€

x 2 − 4 x +12( ) − x 2[ ] dx

0

3

∫

+ x 2 − x 2 − 4 x +12( )[ ] dx3

4

∫

= −4 x +12[ ] dx0

3

∫ + 4x −12[ ] dx3

4

∫

= −2x 2 +12x[ ]3

0+ 2x 2 −12x[ ]

4

3

= −2 32( ) +12 3( )[ ] − −2 02

( ) +12 0( )[ ] + 2 42( ) −12 4( )[ ] − 2 32

( ) −12 3( )[ ]

= –18 + 36 + 0 – 0 + 32 – 48 – 18 + 36

6.4 Example D

Find the area between the curves y = x2 and y = x2 – 4x + 12 on the interval 0 ≤ x ≤ 4.

0

15

50

x = 3

€

x 2 − 4 x +12( ) − x 2[ ] dx

0

3

∫

+ x 2 − x 2 − 4 x +12( )[ ] dx3

4

∫

= −4 x +12[ ] dx0

3

∫ + 4x −12[ ] dx3

4

∫

= −2x 2 +12x[ ]3

0+ 2x 2 −12x[ ]

4

3

= −2 32( ) +12 3( )[ ] − −2 02

( ) +12 0( )[ ] + 2 42( ) −12 4( )[ ] − 2 32

( ) −12 3( )[ ]

= –18 + 36 + 0 – 0 + 32 – 48 – 18 + 36= 20

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

In this case, although no interval is specified, by looking at the graph we can see that there are two points of intersection.

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

In this case, although no interval is specified, by looking at the graph we can see that there are two points of intersection.

Between these two points the area is bounded, and lies between the two curves.

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

In this case, although no interval is specified, by looking at the graph we can see that there are two points of intersection.

Between these two points the area is bounded, and lies between the two curves.

We need to solve for the intersections to find the boundaries (or limits) of integration.

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

We need to solve for the intersections to find the boundaries (or limits) of integration.

842 2 xxx

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

We need to solve for the intersections to find the boundaries (or limits) of integration.

860

8422

2

xx

xxx

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

We need to solve for the intersections to find the boundaries (or limits) of integration.

420

860

8422

2

xx

xx

xxx

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

4,2

420

860

8422

2

x

xx

xx

xxx

x = 4

x = 2

We need to solve for the intersections to find the boundaries (or limits) of integration.

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

x = 4

x = 2

€

2x − x 2 − 4x + 8( )[ ] dx2

4

∫

“above”

“below”

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

x = 4

x = 2

€

2x − x 2 − 4x + 8( )[ ] dx2

4

∫ = −x 2 + 6x − 8[ ] dx2

4

∫

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

x = 4

x = 2

€

2x − x 2 − 4 x + 8( )[ ] dx2

4

∫ = −x 2 + 6x − 8[ ] dx2

4

∫

= −1

3x 3 + 3x 2 − 8x

⎡ ⎣ ⎢

⎤ ⎦ ⎥4

2

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

x = 4

x = 2

€

2x − x 2 − 4x + 8( )[ ] dx2

4

∫ = −x 2 + 6x − 8[ ] dx2

4

∫

= −1

3x 3 + 3x 2 − 8x

⎡ ⎣ ⎢

⎤ ⎦ ⎥4

2

= −1

34( )

3+ 3 4( )

2− 8 4( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥− −

1

32( )

3+ 3 2( )

2− 8 2( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

x = 4

x = 2

€

2x − x 2 − 4 x + 8( )[ ] dx2

4

∫ = −x 2 + 6x − 8[ ] dx2

4

∫

= −1

3x 3 + 3x 2 − 8x

⎡ ⎣ ⎢

⎤ ⎦ ⎥4

2

= −1

34( )

3+ 3 4( )

2− 8 4( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥− −

1

32( )

3+ 3 2( )

2− 8 2( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −64

3+ 48 − 32 +

8

3− 12 + 16

6.4 Example E

Find the area between the curves y = 2x and y = x2 – 4x + 8.

0

10

5

0

x = 4

x = 2

€

2x − x 2 − 4x + 8( )[ ] dx2

4

∫ = −x 2 + 6x − 8[ ] dx2

4

∫

= −1

3x 3 + 3x 2 − 8x

⎡ ⎣ ⎢

⎤ ⎦ ⎥4

2

= −1

34( )

3+ 3 4( )

2− 8 4( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥− −

1

32( )

3+ 3 2( )

2− 8 2( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥

= −64

3+ 48 − 32 +

8

3− 12 + 16

=4

3

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

How many integrals will be needed?

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

How many integrals will be needed? Note that there are three points of intersection.

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

How many integrals will be needed? Note that there are three points of intersection.

A well-placed vertical indicated that two integrals will be sufficient—there are two sub-intervals, each of which has its own “function above” and “function below”.

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

How many integrals will be needed? Note that there are three points of intersection.

A well-placed vertical indicated that two integrals will be sufficient—there are two sub-intervals, each of which has its own “function above” and “function below”.

Finding the intersections will give us the correct boundaries of integration for each integral.

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

Looking at the graph, one of the corners of the bounded area is clearly (0, 0).

x = 0

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

Looking at the graph, one of the corners of the bounded area is clearly (0, 0).

We’ll need to solve for the other two.

x = 0

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

y = 6x and y = 8 – 2x.The “top” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0xx 286

y = 6x and y = 8 – 2x.The “top” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

88

286

x

xx

y = 6x and y = 8 – 2x.The “top” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

1

88

286

x

x

xx x = 0

x = 1y = 6x and y = 8 – 2x.

The “top” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1y = x2 and y = 8 – 2x.

The “right” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

xx 282

y = x2 and y = 8 – 2x.The “right” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

082

282

2

xx

xx

y = x2 and y = 8 – 2x.The “right” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

024

082

282

2

xx

xx

xx

y = x2 and y = 8 – 2x.The “right” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

2,4

024

082

282

2

x

xx

xx

xx

y = x2 and y = 8 – 2x.The “right” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

2,4

024

082

282

2

x

xx

xx

xx

Since –4 is outside our interval of concern, we’ll only be using x = 2.

x = 2y = x2 and y = 8 – 2x.

The “right” corner is the intersection of

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

We’ll need to be careful in setting up the two needed integrals.

x = 0

x = 1

x = 2

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

We’ll need to be careful in setting up the two needed integrals.

For 0 ≤ x ≤ 1, y = 6x is “above” and y = x2 is “below”.

x = 0

x = 1

x = 2

y = 6x

y = x2

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

We’ll need to be careful in setting up the two needed integrals.

For 0 ≤ x ≤ 1, y = 6x is “above” and y = x2 is “below”.

For 1 ≤ x ≤ 2, y = 8 – 2x is “above” and y = x2 is “below”.

x = 0

x = 1

x = 2

y = 6x

y = 8 – 2x

y = x2

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

x = 2

y = 6x

y = 8 – 2x

€

6x − x 2[ ] dx

0

1

∫ + 8 − 2x − x 2[ ] dx

1

2

∫

“above” “above”

“below” “below”

y = x2

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

x = 2

y = 6x

y = 8 – 2x

€

6x − x 2[ ] dx

0

1

∫ + 8 − 2x − x 2[ ] dx

1

2

∫

y = x2

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

x = 2

y = 6x

1

2

3

18

0

1

3

13

286

3232

2

121

02

xxxxx

dxxxdxxx

y = 8 – 2x

Combining like terms is not an option—the two polynomials have different boundaries to evaluate.

y = x2

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

x = 2

y = 6x

€

6x − x 2[ ] dx

0

1

∫ + 8 − 2x − x 2[ ] dx

1

2

∫

= 3x 2 −1

3x 3 ⎡

⎣ ⎢ ⎤ ⎦ ⎥1

0+ 8x − x 2 −

1

3x 3 ⎡

⎣ ⎢ ⎤ ⎦ ⎥2

1

= 3 12( ) −

1

313( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥− 3 02

( ) −1

303

( ) ⎡ ⎣ ⎢

⎤ ⎦ ⎥

+ 8 2( ) − 2( )2

−1

32( )

3 ⎡ ⎣ ⎢

⎤ ⎦ ⎥− 8 1( ) − 1( )

2−

1

31( )

3 ⎡ ⎣ ⎢

⎤ ⎦ ⎥

y = 8 – 2x

y = x2

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

x = 2

y = 6x

y = 8 – 2x

€

6x − x 2[ ] dx

0

1

∫ + 8 − 2x − x 2[ ] dx

1

2

∫

= 3x 2 −1

3x 3 ⎡

⎣ ⎢ ⎤ ⎦ ⎥1

0+ 8x − x 2 −

1

3x 3 ⎡

⎣ ⎢ ⎤ ⎦ ⎥2

1

= 3 12( ) −

1

313( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥− 3 02

( ) −1

303

( ) ⎡ ⎣ ⎢

⎤ ⎦ ⎥

+ 8 2( ) − 2( )2

−1

32( )

3 ⎡ ⎣ ⎢

⎤ ⎦ ⎥− 8 1( ) − 1( )

2−

1

31( )

3 ⎡ ⎣ ⎢

⎤ ⎦ ⎥

= 3 −1

3− 0 + 0 +16 − 4 −

8

3− 8 +1+

1

3

y = x2

6.4 Example F

Find the area bounded by the curves y = x2, y = 6x and y = 8 – 2x on the interval 0 ≤ x ≤ 2.

0

10

5

0

x = 0

x = 1

x = 2

y = 6x

y = 8 – 2x

€

6x − x 2[ ] dx

0

1

∫ + 8 − 2x − x 2[ ] dx

1

2

∫

= 3x 2 −1

3x 3 ⎡

⎣ ⎢ ⎤ ⎦ ⎥1

0+ 8x − x 2 −

1

3x 3 ⎡

⎣ ⎢ ⎤ ⎦ ⎥2

1

= 3 12( ) −

1

313( )

⎡ ⎣ ⎢

⎤ ⎦ ⎥− 3 02

( ) −1

303

( ) ⎡ ⎣ ⎢

⎤ ⎦ ⎥

+ 8 2( ) − 2( )2

−1

32( )

3 ⎡ ⎣ ⎢

⎤ ⎦ ⎥− 8 1( ) − 1( )

2−

1

31( )

3 ⎡ ⎣ ⎢

⎤ ⎦ ⎥

= 3 −1

3− 0 + 0 +16 − 4 −

8

3− 8 +1+

1

3

=16

3

y = x2

6.4 Final Notes

The key to all of these is looking at the graph to determine which function is above the other, finding intersections as needed, then setting up and evaluating the appropriate integrals.

6.4 Final Notes

The key to all of these is looking at the graph to determine which function is above the other, finding intersections as needed, then setting up and evaluating the appropriate integrals.

In the next section we’ll delve into applications which use all of the integration concepts developed so far. If a function represents a rate of change, we’ll use an integral (= area under the curve) to determine an amount.

6.4 Final Notes

The key to all of these is looking at the graph to determine which function is above the other, finding intersections as needed, then setting up and evaluating the appropriate integrals.

In the next section we’ll delve into applications which use all of the integration concepts developed so far. If a function represents a rate of change, we’ll use an integral (= area under the curve) to determine an amount.

For example, from a velocity function, which is the rate of change of position with respect to time, we can integrate to determine the distance traveled.

6.4 Final Notes

The key to all of these is looking at the graph to determine which function is above the other, finding intersections as needed, then setting up and evaluating the appropriate integrals.

In the next section we’ll delve into applications which use all of the integration concepts developed so far. If a function represents a rate of change, we’ll use an integral (= area under the curve) to determine an amount.

For example, from a velocity function, which is the rate of change of position with respect to time, we can integrate to determine the distance traveled.

Another example: Knowing how costs are changing with respect to the number of units produced (Marginal Cost) we can integrate to find the amount of Cost.