1. More on Areas

2. We are to find the area of a given enclosed region R.
R
More on Areas

3. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
R
More on Areas
x

4. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
R
a=x b=x
x

5. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
R
a=x b=x
x

6. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
R
L(x)
a=x b=x
x
x

7. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b]
R
a=x b=x
x
L(x)
x

8. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b]
x1 x2 xi–1 xi
R
a=x0 b=xn
x

9. More on Areas
We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x, x, x, .. x=b} be a regular partition of
012n[a, b] and select an arbitrary point x*
in each sub-interval
i [x , x].
i–1ix1 x2 xi–1 xi
R
a=x0 b=xn
x

10. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x, x, x, .. x=b} be a regular partition of
012n[a, b] and select an arbitrary point x*
in each sub-interval
i [x , x].
i–1ia=xxxb=x
0 1 2
x1 *
xi–1 xi
R
More on Areas
x

11. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi].
x1 * x2 *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
*
x

12. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi].
x1 * x2 * x3 *
xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
*
x

13. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi].
x1 * x2 * x3 *
xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
L(xi)
*
*
x

14. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi].
x1 * x2 * x3 *
xi *
cross–sectional
length at xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
L(xi)
*
*
x

15. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi]. Let Δx = the width of each subinterval.
x1 * x2 * x3 *
xi *
cross–sectional
length at xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
*
L(xi*)
x

16. We are to find the area of a given enclosed region R.
Take a ruler x and measure R from one end to the other end,
and assume that R spans from x = a to x = b.
Let L(x) = cross–sectional length at a generic x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b] and select an arbitrary point xi in each sub-interval
[x i–1, xi]. Let Δx = the width of each subinterval. The rectangle
with L(xi) as height and Δx as width approximates the area in R
that is spanned from xi–1 to xi.
x1 * x2 * x3 *
xi *
cross–sectional
length at xi *
a=xxxb=x
0 1 2
xi–1 xi
R
More on Areas
*
*
Δx
L(xi*)
x

17. More on Areas
R
x1 * x2 * x3 * xi *
a=x0 x b=x 1 x2 xi–1 xi
The Riemann sum
x

18. More on Areas
R
Δx
L(x1*)
x1 * x2 * x3 * xi *
a=x0 x b=x 1 x2 xi–1 xi
L(x1The Riemann sum *)Δx
x

19. More on Areas
R
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
a=x0 x b=x 1 x2 xi–1 xi
L(x1*)Δx+ L(x2The Riemann sum *)Δx+ …
x

20. More on Areas
Δx
*
L(xi)
R
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
Δx
L(xn* )
xn* x
a=x0 x b=x 1 x2 xi–1 xi
L(x1*)Δx+ L(x2*)Δx+ … L(xnThe Riemann sum *)Δx

21. More on Areas
Δx
L(xi)
R
x1 * x2 * x3 * xi *
Δx
L(xn* )
a=x0 x b=x 1 x2 xi–1 xi
n
The Riemann sum L(x*)Δx+
L(x *)Δx+ … L(x*)Δx =
Σ L(x*)Δx 12nii=1
of all such rectangles approximates the area of R.
L(x1) L(x2* *)
*
Δx
Δx
xn*
x

22. More on Areas
Δx
*
L(xi)
R
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
Δx
L(xn* )
xn*
a=x0 x b=x 1 x2 xi–1 xi
n
The Riemann sum L(x*)Δx+
L(x *)Δx+ … L(x*)Δx =
Σ L(x*)Δx 12nii=1
of all such rectangles approximates the area of R.
In fact the mathematical definition of the area of R is the limit
n
Σ L(xiof the Riemann sums *)Δx as Δx 0 or n ∞,
x

23. More on Areas
Δx
*
L(xi)
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
Δx
L(xn* )
xn*
a=x0 x b=x 1 x2 xi–1 xi
n
Σ L(xiL(x *)Δx 2*)Δx+ … L(xnThe Riemann sum *)Δx =
of all such rectangles approximates the area of R.
In fact the mathematical definition of the area of R is the limit
the area of R is A = ∫
b
L(x) dx.
x=a
R
L(x1*)Δx+
i=1
n
Σ L(xiof the Riemann sums *)Δx as Δx 0 or n ∞, and
by the FTC
x

24. More on Areas
Δx
*
L(xi)
Δx
Δx
L(x1) L(x2* *)
x1 * x2 * x3 * xi *
Δx
L(xn* )
xn*
a=x0 x b=x 1 x2 xi–1 xi
x
n
Σ L(xiL(x *)Δx 2*)Δx+ … L(xnThe Riemann sum *)Δx =
of all such rectangles approximates the area of R.
In fact the mathematical definition of the area of R is the limit
the area of R is A = ∫
b
L(x) dx.
x=a
R
L(x1*)Δx+
i=1
n
Σ L(xiof the Riemann sums *)Δx as Δx 0 or n ∞, and
by the FTC
Theorem. The area of a 2D region is the definite integral of its
cross–section (length) function.

25. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2

26. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
y = x2
y = –x2 + 2x
x

27. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area.
y = x2
y = –x2 + 2x
x

28. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
x

29. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x

30. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x x

31. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
x

32. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
x

33. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
L(x) =(–x2 + 2x) – x2
x x
L(x) =(–x2 + 2x) – x2

34. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
1
(–x2 + 2x) – x2 dx =
Hence the area is ∫
x=0
L(x) =(–x2 + 2x) – x2
x x
L(x) =(–x2 + 2x) – x2

35. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
1
(–x2 + 2x) – x2 dx = ∫
Hence the area is ∫
x=0
L(x) =(–x2 + 2x) – x2
x x
1
–2x2 + 2x dx
0
L(x) =(–x2 + 2x) – x2

36. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
1
(–x2 + 2x) – x2 dx = ∫
Hence the area is ∫
x=0
1
–2x2 + 2x dx
=
–2
3
1
x3 + x2 |
0
0
L(x) =(–x2 + 2x) – x2
x x
L(x) =(–x2 + 2x) – x2

37. More on Areas
Example A.
a. Find the area bounded by y = –x2 + 2x and y = x2
We need to find the span
of the area. They are the
x–coordinates of the
intersections of the curves.
y = x2
y = –x2 + 2x
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
1
(–x2 + 2x) – x2 dx = ∫
Hence the area is ∫
x=0
1
–2x2 + 2x dx
=
–2
3
1
=
x3 + x2 |
0
0
1
3
L(x) =(–x2 + 2x) – x2
x x
L(x) =(–x2 + 2x) – x2

38. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

39. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
f(x) = 2x – x3
g(x) = –x2

40. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts.
f(x) = 2x – x3
g(x) = –x2

41. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x).
f(x) = 2x – x3
g(x) = –x2

42. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
f(x) = 2x – x3
g(x) = –x2

43. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2

44. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3

45. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0

46. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0

47. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.

48. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.
Therefore the span for the 1st area is from x = –1 to x = 0

49. More on Areas
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
The bounded area consists
of two parts. The first part
is bounded above by g(x)
and below by f(x). The
second part is f(x) on top.
We set the equations equal
to solve for the span.
f(x) = 2x – x3
g(x) = –x2
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.
Therefore the span for the 1st area is from x = –1 to x = 0
and the span for the 2nd area is from x = 0 to x = 2.

50. 0
–x2 – (2x – x3) dx
∫
x= –1
More on Areas
f(x) = x – x3
2
2x – x3 – (–x2) dx
g(x) = –x2
Hence the total bounded area is
+ ∫
x= 0
–1 0 2

51. 0
–x2 – (2x – x3) dx
∫
x= –1
More on Areas
f(x) = x – x3
2
2x – x3 – (–x2) dx
g(x) = –x2
Hence the total bounded area is
+ ∫
x= 0
=
–1 0 2
0
x3 – x2 – 2x dx + ∫0
∫–1
2
– x3 + x2 + 2x dx

52. 0
–x2 – (2x – x3) dx
∫
x= –1
More on Areas
f(x) = x – x3
2
2x – x3 – (–x2) dx
2
g(x) = –x2
Hence the total bounded area is
+ ∫
x= 0
=
–1 0 2
0
x3 – x2 – 2x dx + ∫0
∫–1
2
– x3 + x2 + 2x dx
x4
4
= –
x3
3
– x2 |
0
–1
+
–x4
4
+ x3
3
+ x2 |
0

53. 0
–x2 – (2x – x3) dx
∫
x= –1
More on Areas
f(x) = x – x3
2
2x – x3 – (–x2) dx
2
g(x) = –x2
Hence the total bounded area is
+ ∫
x= 0
=
–1 0 2
0
x3 – x2 – 2x dx + ∫0
∫–1
2
– x3 + x2 + 2x dx
x4
4
= –
x3
3
– x2 |
0
–1
+
–x4
4
+ x3
3
+ x2 |
0
=
5
12
+
8
3
=
37
12

54. 0
–x2 – (2x – x3) dx
∫
x= –1
More on Areas
f(x) = x – x3
2
2x – x3 – (–x2) dx
2
g(x) = –x2
Hence the total bounded area is
+ ∫
x= 0
=
–1 0 2
0
x3 – x2 – 2x dx + ∫0
∫–1
2
– x3 + x2 + 2x dx
x4
4
= –
x3
3
– x2 |
0
–1
+
–x4
4
+ x3
3
+ x2 |
0
=
5
12
+
8
3
=
37
12
A type I region R is a region that
is bounded on top by a curve
y = f(x) and bounded below by
y = g(x) from x = a to x = b.
The above examples are regions
of type I.

55. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.

56. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.
y = b
y = a
x = g(y) x = f(y)

57. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.
Its cross–sectional length is L(y) = f(y) – g(y).
y = b
y = a
x = g(y) x = f(y)

58. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.
Its cross–sectional length is L(y) = f(y) – g(y).
y = b
y = a
x = g(y) x = f(y)
L(y) = f(y) – g(y)

59. More on Areas
A type II region R is a region that is bounded to the right by a
curve x = f(y) and to the left by x = g(y) from y = a to y = b.
Its cross–sectional length is L(y) = f(y) – g(y).
b
Therefore the area of R = ∫ f(y) – g(y) dy
y = b
y = a
y=a
x = g(y) x = f(y)
L(y) = f(y) – g(y)

60. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.

61. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
y = x – 2
y = √x

62. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question.
y = x – 2
y = √x

63. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x

64. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4

65. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4

66. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.

67. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.

68. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
4

69. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4.
4

70. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
4
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4. But the
lower boundary is not a single function.

71. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
4
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4. But the
lower boundary is not a single function.

72. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
y = x – 2
y = √x
4
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4. But the
lower boundary is not a single function. Therefore to find the
area of the region, we have to split it into two pieces,
I and II as shown.

73. More on Areas
Example B. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
I II
y = x – 2
y = √x
4
x = x2 – 4x + 4
0 = x2 – 5x + 4
0 = (x – 4)(x – 1) so x = 1 and 4.
However x = 4 is the only good solution.
As a type I region, the span is from x = 0 to x = 4. But the
lower boundary is not a single function. Therefore to find the
area of the region, we have to split it into two pieces,
I and II as shown.

74. More on Areas
I II
y = x – 2
y = √x
2 4
The total area is (area I) + (area II) i.e.

75. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2 4

76. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
4

77. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
2
| +
0
4

78. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
4

79. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
=
2(2)3/2
3
+
4

80. More on Areas
I II
y = x – 2
y = √x
4
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
=
2(2)3/2
3
+
2(4)3/2
3
[( –
42
2
+ 8 ) –
2(2)3/2
–
3
22
2
( + 4 ) ]

81. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
=
2(2)3/2
3
+
2(4)3/2
3
[( –
42
2
+ 8 ) –
2(2)3/2
–
3
22
2
( + 4 ) ]
=
2(2)3/2
+
3
16
3
–
2(2)3/2
3
– 2
4

82. More on Areas
I II
y = x – 2
y = √x
The total area is (area I) + (area II) i.e.
2
√x dx +
∫
x= 0
2
4
√x – (x – 2) dx
∫
x= 2
=
2x3/2
3
+
2x3/2
3
( –
x2
2
2
| + 2x )
0
4
|
2
=
2(2)3/2
3
+
2(4)3/2
3
[( –
42
2
+ 8 ) –
2(2)3/2
–
3
22
2
( + 4 ) ]
=
2(2)3/2
+
3
16
3
–
2(2)3/2
3
– 2 =
10
3
4

83. More on Areas
y = x – 2
y = √x
However, we may view this as a type II region.

84. More on Areas
y = x – 2
y = √x
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2.

85. More on Areas
y = x – 2
y = √x
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.

86. More on Areas
y = x – 2
y = √x
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2

87. More on Areas
y = x – 2
y = √x
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2

88. More on Areas
y = 2
y = √x
so x = y2
y = x – 2 so x = y + 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2

89. More on Areas
y = 2
y = √x
so x = y2
y = x – 2 so x = y + 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2

90. More on Areas
L(y) = y + 2 – y2
y = x – 2 so x = y + 2
y = √x
so x = y2
y = 2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.

91. More on Areas
L(y) = y + 2 – y2
y = x – 2 so x = y + 2
y = √x
so x = y2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.
Therefore the area is
2
∫ y + 2 – y2 dy
y = 0
y = 2

92. More on Areas
L(y) = y + 2 – y2
y = x – 2 so x = y + 2
y = √x
so x = y2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.
Therefore the area is
2
∫ y + 2 – y2 dy
y = 0
y2
2
= +
2y –
y3
|
2 3 0
y = 2

93. More on Areas
L(y) = y + 2 – y2
y = x – 2 so x = y + 2
y = √x
so x = y2
However, we may view this as a type II region. It spans
from y = 0 to y = 2. Solve for x for the boundary functions.
The right boundary is x = f(y) = y + 2 and the left boundary
is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.
Therefore the area is
2
∫ y + 2 – y2 dy
y = 0
y2
2
= +
2y –
y3
|
2 3 0
= 6 –
8
3
=
10
3
y = 2