calculus with algebra and trigonometry ii lecture 15 areas
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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