Calculus with Algebra and Trigonometry IILecture 15

Areas by integration

Mar 12, 2015

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 1 / 21

Example 1

Find the area bounded by the lines x = ±1 and the function

y =1

1 + x2

Area =

∫ 1

−1

1

1 + x2dx =

[tan−1 x

]1−1

= tan−1(1)− tan−1(−1)

=π

4−(−π

4

)=π

2

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 2 / 21

Example 2

Find the area bounded by the x axis, the line x = 3 and the function

y = x ln x

Area =

∫ 3

1x ln x dx =

[!

2x2 ln x

]31

−∫ 3

1

1

2x2(

1

x

)dx

=9

2ln 3−

[1

4x2]31

=9

2ln 3− 9

4−(−1

4

)=

9

2ln 3− 2

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 3 / 21

Example 3

Find the area bounded the line y = 12 and the function

y = sin x 0 ≤ x ≤ π

To find the limits for the integration we need to find the intersection points

sin x =1

2⇒ x =

π

6,

5π

6

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 4 / 21

The area under the sin x graph is

A1 =

∫ 5π6

π6

sin x dx

= [− cos x ]5π6π6

= − cos

(5π

6

)−(− cos

(π6

))= −

(−√

3

2

)−

(−√

3

2

)= 2 = sqrt3

The area below y = 12 is

A2 =

∫ 5π6

π6

1

2dx =

1

2

(5π

6− π

6

)=π

3

Area = A1 − A2 =√

3− π

3

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 5 / 21

Example 4

Find the area enclosed by the curve

y2 = x2 − x4

Using the symmetry of the graph, we are going to calculate the area in thefirst quadrant and multiply by 4. The area in the first quadrant is given by

A1 =

∫ 1

0x√

1− x2 dx

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 6 / 21

To evaluate the integral we will use u substitution. Let

u = 1− x2 ⇒ du = −2x dx ⇒ dx = −du

2x

x = 0 ⇒ u = 1 x = 1 ⇒ u = 0

Area =

∫ 1

0x√

1− x2 dx

=

∫ 0

1x√u

(−du

2x

)= −1

2

∫ 0

1

√u du

= −1

2

[2

3u3/2

]01

=1

3

So the area enclosed is 43 .

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 7 / 21

Example 5

Find ∫ 2

−1|x3 − x2 − 2x | dx

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 8 / 21

Example 5

Find ∫ 2

−1|x3 − x2 − 2x | dx

The function f (x) = x3 − x2 − 2x has zeros at x = −1, 0, 2. It is positivebetween -1 and zero and negative between 0 and 2.

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 9 / 21

Thus∫ 2

−1|x3 − x2 − 2x | dx =

∫ 0

−1(x3 − x2 − 2x) dx −

∫ 2

0(x3 − x2 − 2x) dx

=

[x4

4− x3

3− x2

]0−1

−[x4

4− x3

3− x2

]20

= −(

1

4+

1

3− 1

)−(

4− 8

3− 4

)=

37

12

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 10 / 21

Area between two functions

To calculate the area between two functions, imagine the area divided intoskinny rectangles as shown below. At a given vlaue of x the rectangle willhave height = f (x)− g(x) and width dx . The area will be given by addingup all the rectangles i.e.

Area =

∫ x(B)

x(A)(f (x)− g(x)) dx

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 11 / 21

Example 6

Find the area bounded by the y axis and the functions y = sin x andy = cos x

To find the bounds for the integration we need to find the intersectionpoint.

sin x = cos x ⇒ x =π

4

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 12 / 21

The area is then given by

Area =

∫ π4

0(cos x − sin x) dx

= [sin x + cos x ]π40

=

(√2

2+

√2

2

)− 1

=√

2− 1

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 13 / 21

Example 7

Find the area bounded by the functions y = ex/2, y = e−x/2 and the linex = 2 ln 2.

Area =

∫ 2 ln 2

0(ex/2 − e−x/2) dx

= [2ex/2 + 2e−x/2]2 ln 20

=

(2(2) + 2× 1

2

)− 4 = 1

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 14 / 21

Example 8

Find the area bounded by the function y = 1 + cos x and the linesx = π, y = 2.

Area =

∫ π

0(2− (1 + cos x) dx

= [x − sin x ]π0

= π

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 15 / 21

Example 9

Find the area between the parabolas

THe intersection points are given by

6x − x2 = x2 − 2x ⇒ 8x − 2x2 = 0 ⇒ x = 0, 4

Area =

∫ 4

0(6x − x2 − (x2 − 2x)) dx

=

∫ 4

0(8x − 2x2) dx

=

[4x2 − 2

3x3]40

=64

3

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 16 / 21

Integrating in the y direction

Sometimes it is more convenient to integrate in the y direction If thefunctions are given in the form x = f (y) and x = g(y) then the areabetween them is given by

Area =

∫ y(A)

y(B)(f (y)− g(y)) dy

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 17 / 21

Example 10

Find the area in the first quadrant beween the y axis and the curvex = −y3 + 2y2 + 3y .

Area =

∫ 2

0(−y3 + 2y2 + 3y) dy

=

[−y4

4+

2

3y3 +

3

2y2]20

=

(−4 +

16

3+ 6

)=

22

3

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 18 / 21

Example 11

Find the area between the parabola y2 = 4x and the line y = 2x − 4.

The intersection points are given by

y2 = 2(y + 4) ⇒ y2 − 2y − 8 = 0 ⇒ y = −2, 4

Area =

∫ 4

−2

(y + 4

2− y2

4

)) dy

=

[(y + 4)2

4− y3

12

]4−2

=

(16− 16

3

)−(

1 +2

3

)= 6

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 19 / 21

Example 12

Find the area between the parabolas x = 2(y − 1)2 and x = 1 + (y − 1)2.

The interesction points are given by

x = 2(y − 1)2 = 2(x − 1) ⇒ x = 2 ⇒ y = 0, 2

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 20 / 21

The area is then given by

Area =

∫ 2

0(1 + (y − 1)2 − 2(y − 1)2) dy

=

∫ 2

0(1− (y − 1)2) dy

=

∫ 2

0(2y − y2) dy

=

[y2 − y3

3

]20

=

(4− 8

3

)=

4

3

Calculus with Algebra and Trigonometry II Lecture 15Areas by integrationMar 12, 2015 21 / 21