Fundamental Theorem of Calculus and Signed Areas

1. More on Area

2. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
a=x b=x
R
More on Area

3. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
a=x b=x
R
More on Area
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.

4. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
a=x b=x
R
More on Area
L(x)
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.

5. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
a=x b=x
R
More on Area
L(x)
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b]

6. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.
a=x0 b=xn
Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of
[a, b]
x1 x2 xi–1 xi
R
More on Area
L(x)

7. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.
a=x0 b=xn
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 xi–1 xi
R
More on Area
L(x)
*

8. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.
a=x0 b=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
x1
*
xi–1 xi
R
More on Area
L(x)

9. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.
a=x0 b=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
x1
* x2
*
xi–1 xi
R
More on Area
L(x)
*

10. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.
a=x0 b=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
x1
* x2
* x3
*
xi–1 xi
xi
*
R
More on Area
L(xi)
*
*

11. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.
a=x0 b=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
x1
* x2
* x3
*
xi–1 xi
xi
*
R
More on Area
*
L(xi)*

12. Given a 2D region R, and we would like to find its area.
Take a ruler and measure R from one end to the another, let x
be the measurements and that R spans from x = a to x = b.
Let L(x) = cross–sectional length at x.
a=x0 b=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
x1
* x2
* x3
*
xi–1 xi
xi
*
R
More on Area
*
*
Δx
L(xi)*

13. More on Area
a=x0 b=xx1 x2 xi–1 xi
x1
* x2
* x3
* xi
*
R
The Riemann sum

14. More on Area
a=x0 b=xx1 x2 xi–1 xi
x1
* x2
* x3
* xi
*
R
L(x1)Δx*The Riemann sum
L(x1)*
Δx

15. More on Area
a=x0 b=xx1 x2 xi–1 xi
x1
* x2
* x3
* xi
*
R
L(x1)Δx+* L(x2)Δx+ …*The Riemann sum
L(x1) L(x2)* *
Δx
Δx

16. More on Area
a=x0 b=xx1 x2 xi–1 xi
x1
* x2
* x3
* xi
*
L(xi)
Δx
R
L(x1)Δx+* L(x2)Δx+ …* L(xn)Δx*The Riemann sum
L(x1) L(x2)* *
*
Δx
Δx
L(xn)*
Δx
xn
*

17. More on Area
a=x0 b=xx1 x2 xi–1 xi
x1
* x2
* x3
* xi
*
L(xi)
Δx
R
L(x1)Δx+*
i=1
n
∑ L(xi)Δx*L(x2)Δx+ …* L(xn)Δx =*The Riemann sum
of all such rectangles approximates the area of R.
L(x1) L(x2)* *
*
Δx
Δx
xn
*
Δx
L(xn)*

18. More on Area
a=x0 b=xx1 x2 xi–1 xi
x1
* x2
* x3
* xi
*
L(xi)
Δx
In fact the mathematical definition of the area of R is the limit
R
L(x1)Δx+*
i=1
n
∑ L(xi)Δx*L(x2)Δx+ …* L(xn)Δx =*The Riemann sum
of all such rectangles approximates the area of R.
n
∑ L(xi)Δx as Δx 0 or n ∞,*of the Riemann sums
L(x1) L(x2)* *
*
Δx
Δx
xn
*
Δx
L(xn)*

19. More on Area
a=x0 b=xx1 x2 xi–1 xi
x1
* x2
* x3
* xi
*
L(xi)
Δx
In fact the mathematical definition of the area of R is the limit
the area of R =∫x=a
b
L(x) dx.
R
L(x1)Δx+*
i=1
n
∑ L(xi)Δx*L(x2)Δx+ …* L(xn)Δx =*The Riemann sum
of all such rectangles approximates the area of R.
n
∑ L(xi)Δx as Δx 0 or n ∞, so*of the Riemann sum
by the FTC
L(x1) L(x2)* *
*
Δx
Δx
xn
*
Δx
L(xn)*

20. More on Area
a=x0 b=xx1 x2 xi–1 xi
x1
* x2
* x3
* xi
*
L(xi)
Δx
In fact the mathematical definition of the area of R is the limit
the area of R =∫x=a
b
L(x) dx.
R
L(x1)Δx+*
i=1
n
∑ L(xi)Δx*L(x2)Δx+ …* L(xn)Δx =*The Riemann sum
of all such rectangles approximates the area of R.
n
∑ L(xi)Δx as Δx 0 or n ∞, so*of the Riemann sum
by the FTC
Theorem. The area of a 2D region is the definite integral of its
cross–section (length) function.
L(x1) L(x2)* *
*
Δx
Δx
xn
*
Δx
L(xn)*

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

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

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

24. 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.
More on Area
y = –x2 + 2x
y = x2

25. 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.
More on Area
y = –x2 + 2x
y = x2
Set the equations equal to
find the intersection points.

26. 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.
More on Area
y = –x2 + 2x
y = x2
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x

27. 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.
More on Area
y = –x2 + 2x
y = x2
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x

28. 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.
More on Area
y = –x2 + 2x
y = x2
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.

29. 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.
More on Area
y = –x2 + 2x
y = x2
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
Hence the area is ∫x=0
1
(–x2 + 2x) – x2 dx

30. 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.
More on Area
y = –x2 + 2x
y = x2
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
Hence the area is ∫x=0
1
(–x2 + 2x) – x2 dx = ∫
1
–2x2 + 2x dx
0

31. 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.
More on Area
y = –x2 + 2x
y = x2
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
Hence the area is ∫x=0
1
(–x2 + 2x) – x2 dx = ∫
1
–2x2 + 2x dx
=
–2
3
x3 + x2 |
0
0
1

32. 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.
More on Area
y = –x2 + 2x
y = x2
Set the equations equal to
find the intersection points.
x2 = –x2 + 2x
2x2 = 2x
2x(x – 1) = 0 so x = 0, 1.
Hence the area is ∫x=0
1
(–x2 + 2x) – x2 dx = ∫
1
–2x2 + 2x dx
=
–2
3
x3 + x2 |
0
0
1
=
1
3

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

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

35. 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).
More on Area
f(x) = 2x – x3
g(x) = –x2

36. 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 has f(x) on
top.
More on Area
f(x) = 2x – x3
g(x) = –x2

37. 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 has f(x) on
top. We set the equations
equal to solve for the span.
More on Area
f(x) = 2x – x3
g(x) = –x2

38. 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 has f(x) on
top. We set the equations
equal to solve for the span.
More on Area
–x2 = 2x – x3
f(x) = 2x – x3
g(x) = –x2

39. 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 has f(x) on
top. We set the equations
equal to solve for the span.
More on Area
–x2 = 2x – x3
x3 – x2 – 2x = 0
f(x) = 2x – x3
g(x) = –x2

40. 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 has f(x) on
top. We set the equations
equal to solve for the span.
More on Area
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
f(x) = 2x – x3
g(x) = –x2

41. 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 has f(x) on
top. We set the equations
equal to solve for the span.
More on Area
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
f(x) = 2x – x3
g(x) = –x2
x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.
–1 0 2

42. 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 has f(x) on
top. We set the equations
equal to solve for the span.
More on Area
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
Therefore the span for the 1st area is from x = –1 to x = 0
f(x) = 2x – x3
g(x) = –x2
x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.
–1 0 2

43. 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 has f(x) on
top. We set the equations
equal to solve for the span.
More on Area
–x2 = 2x – x3
x3 – x2 – 2x = 0
x(x2 – x – 2) = 0
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.
f(x) = 2x – x3
g(x) = –x2
x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.
–1 0 2

44. 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 has f(x) on
top. We set the equations
equal to solve for the span.
More on Area
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.
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
–1 0 2

45. 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 has f(x) on
top. We set the equations
equal to solve for the span.
More on Area
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.
We label them as A1 and A2 as shown.
A1
A2
b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2
–1 0 2

46. More on Area
∫x= –1
0
–x2 – (2x – x3) dx
Hence the total bounded area is
A1
f(x) = x – x3
g(x) = –x2
–1 0 2
A2
A1

47. More on Area
∫x= –1
0
–x2 – (2x – x3) dx
Hence the total bounded area is
+ ∫x= 0
2
2x – x3 – (–x2) dx
A1 A2
f(x) = x – x3
g(x) = –x2
–1 0 2
A2
A1

48. More on Area
= ∫–1
x3 – x2 – 2x dx +∫0
– x3 + x2 + 2x dx
∫x= –1
0
–x2 – (2x – x3) dx
Hence the total bounded area is
+ ∫x= 0
2
2x – x3 – (–x2) dx
0 2
A1 A2
f(x) = x – x3
g(x) = –x2
–1 0 2
A2
A1

49. More on Area
= ∫–1
x3 – x2 – 2x dx +∫0
– x3 + x2 + 2x dx
x4
4
–=
x3
3
– x2|–1
0
+ –x4
4
+ x3
3
+ x2
|
2
∫x= –1
0
–x2 – (2x – x3) dx
Hence the total bounded area is
+ ∫x= 0
2
2x – x3 – (–x2) dx
0 2
A1 A2
f(x) = x – x3
g(x) = –x2
–1 0 2
A2
A1
0

50. More on Area
= ∫–1
x3 – x2 – 2x dx +∫0
– x3 + x2 + 2x dx
x4
4
–=
x3
3
– x2|–1
0
+ –x4
4
+ x3
3
+ x2
|0
2
=
5
12 +
8
3 = 37
12
f(x) = x – x3
g(x) = –x2
–1 0 2
A2
A1
∫x= –1
0
–x2 – (2x – x3) dx
Hence the total bounded area is
+ ∫x= 0
2
2x – x3 – (–x2) dx
0 2
A1 A2

51. More on Area
= ∫–1
x3 – x2 – 2x dx +∫0
– x3 + x2 + 2x dx
x4
4
–=
x3
3
– x2|–1
0
+ –x4
4
+ x3
3
+ x2
|0
2
=
5
12 +
8
3 = 37
12
f(x) = x – x3
g(x) = –x2
–1 0 2
A2
A1
∫x= –1
0
–x2 – (2x – x3) dx
Hence the total bounded area is
+ ∫x= 0
2
2x – x3 – (–x2) dx
0 2
A1 A2
Type I and Type II Regions

52. More on Area
= ∫–1
x3 – x2 – 2x dx +∫0
– x3 + x2 + 2x dx
x4
4
–=
x3
3
– x2|–1
0
+ –x4
4
+ x3
3
+ x2
|0
2
=
5
12 +
8
3 = 37
12
f(x) = x – x3
g(x) = –x2
–1 0 2
A2
A1
∫x= –1
0
–x2 – (2x – x3) dx
Hence the total bounded area is
+ ∫x= 0
2
2x – x3 – (–x2) dx
0 2
A1 A2
Type I and Type II Regions
Area of a region is the integral of
its cross-sectional lengths.

53. More on Area
= ∫–1
x3 – x2 – 2x dx +∫0
– x3 + x2 + 2x dx
x4
4
–=
x3
3
– x2|–1
0
+ –x4
4
+ x3
3
+ x2
|0
2
=
5
12 +
8
3 = 37
12
f(x) = x – x3
g(x) = –x2
–1 0 2
A2
A1
∫x= –1
0
–x2 – (2x – x3) dx
Hence the total bounded area is
+ ∫x= 0
2
2x – x3 – (–x2) dx
0 2
A1 A2
Type I and Type II Regions
Area of a region is the integral of
its cross-sectional lengths. Above examples
are integrals across vertical cross-sectional length L(x).

54. More on Area
= ∫–1
x3 – x2 – 2x dx +∫0
– x3 + x2 + 2x dx
x4
4
–=
x3
3
– x2|–1
0
+ –x4
4
+ x3
3
+ x2
|0
2
=
5
12 +
8
3 = 37
12
f(x) = x – x3
g(x) = –x2
–1 0 2
A2
A1
∫x= –1
0
–x2 – (2x – x3) dx
Hence the total bounded area is
+ ∫x= 0
2
2x – x3 – (–x2) dx
0 2
A1 A2
Type I and Type II Regions
Area of a region is the integral of
its cross-sectional lengths. Above examples
are integrals across vertical cross-sectional length L(x).
We get the same area if the cross-sectional lengths are
horizontal. Let’s distinguish following two types of regions.

55. The basic geometry here is
that for vertical distance y,
we need to find y = f(x),
i.e. y as a function in x.
For horizontal distances,
we need to find x = g(y),
i.e. x as a function in y.
More on Area
y
x
( x= g(y), y = f(x) )
x = g(y)
y = f(x)
x
y
Example B. Given 2x – y2 = –4 for 0 ≤ y,
express y as f(x), and x as g(y). Draw.
Solving for the vertical length y,
2x – y2 = –4 so y2 = 2x + 4,
since 0 ≤ y → y = f(x) = √2x + 4.
(x, y)
2x – y2 = 4
y = √2x + 4
Solving for the horizontal length x,
2x – y2 = –4 so 2x = –4 + y2,
or that x = g(y) = (–4 + y2)/2.
x = (–4 + y2)/2

56. Type I Region
A type I region R is a region that´s
in a vertical stripe starting from
x = a ending at x = b,
bounded on top by y = f(x)
and below by y = g(x) as shown.
It’s vertical cross-sectional length is
L(x) = f(x) – g(x) for a ≤ x ≤ b.
More on Area
x = a x = b
y = f(x)
y = g(x)
L(x) = f(x) – g(x)
The area of a type I region is
the integral of its vertical
cross-sectional length:
The region in example A is viewed as
type I and its area is the integral of
“vertical lines”:
x = 1x = 0
∫
b
L(x) dx
a
=
∫
b
f(x) – g(x) dx
a
∫x=0
1
(–x2 + 2x) – x2 dx =
1
3

57. More on Area
Type Il Region
A type Il region R is a region that´s
in a horizontal stripe starting from
y = c ending at y = d,
bounded to the right by x = f(y)
and to the left by x = g(y) as shown.
It’s horizontal cross-sectional length
is L(y) = f(y) – g(y) for c ≤ y ≤ d.
L(y) = f(y) – g(y)
y = c
y = dx = f(y)x = g(y)
The area of a type II region is
the integral of its horizontal
cross-sectional length:
x = 1x = 0
∫
d
L(y) dy
c
=
∫
d
f(y) – g(y) dy
c
The area in example A is both
types I and II.
But it’s more work to view it as type II
because of the algebra (we would need
to solve for x in terms of y.)
L(y) = ?
y

58. More on Area
Example C. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.

59. More on Area
Example C. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
y = x – 2
y = √x

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

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

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

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

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

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

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

68. More on Area
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
Example C. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
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. But the
lower boundary is not a single function.
4

69. More on Area
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
Example C. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
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. But the
lower boundary is not a single function.
4

70. More on Area
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
Example C. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
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. 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.
4

71. More on Area
The shaded region is the
area in question. For the
span of the region, set
√x = x – 2
Example C. Find the area that is bounded by the positive
x–axis, y = x – 2 and y = √x.
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. 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.
I II
4

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

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

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

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

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

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

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

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

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

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

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

83. More on Area
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.
y = x – 2
y = √x
y = 2

84. More on Area
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
y = x – 2
y = √x
y = 2

85. More on Area
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
y = x – 2
y = √x
y = 2

86. More on Area
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
y = 2
y = √x
so x = y2
y = x – 2 so x = y + 2

87. More on Area
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
y = 2
y = √x
so x = y2
y = x – 2 so x = y + 2

88. More on Area
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.
y = x – 2 so x = y + 2
y = √x
so x = y2
L(y) = y + 2 – y2
y = 2

89. More on Area
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.
y = x – 2 so x = y + 2
y = √x
so x = y2
L(y) = y + 2 – y2
Therefore the area is
y + 2 – y2 dy∫y = 0
2
y = 2

90. More on Area
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.
y = x – 2 so x = y + 2
y = √x
so x = y2
L(y) = y + 2 – y2
Therefore the area is
y + 2 – y2 dy∫y = 0
2
= +
y2
2
2y –
y3
3 0
2
|
y = 2

91. More on Area
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.
y = x – 2 so x = y + 2
y = √x
so x = y2
L(y) = y + 2 – y2
Therefore the area is
y + 2 – y2 dy∫y = 0
2
= +
y2
2
2y –
y3
3 0
2
| = 6 –
8
3
=
10
3
y = 2
The designation of type I and II is mainly a reminder
to compare the integration in different geometry
which may yield simpler solutions.