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What’s the definite integral used for?
f
g

Area of region
between f and g

f

=

=

g

-

Area of region
under f(x)

b

a

a

g

b

b

(x x
∫[f(x)−g )]d =∫ f ( x)dx

f

−

Area of region
under g(x)

∫ g ( x)dx
a
7.1 Areas Between Curves
To find the area:
• divide the area into n strips of
equal width
• approximate the ith strip by a
rectangle with base Δx and height
f(xi) – g(xi).
• the sum of the rectangle areas is a
good approximation
• the approximation is getting better
as n→∞.

y = f(x)

y = g(x)

The area A of the region bounded by the curves y=f(x), y=g(x), and
the lines x=a, x=b, where f and g are continuous and f(x) ≥ g(x) for
all x in [a,b], is
b

A = ∫ [ f ( x) − g ( x)]dx
a
To find the area between 2 curves
(along the x-axis)
• Sketch a graph (if you can)
• Draw a representative rectangle to determine
the upper and lower curves.
• Use the formula:
Area = Top curve – bottom curve
b

A ∫[f(x (x ] x =
=
)−g ) d
a
Ex. Find the area of the region bounded by the graphs
of f(x) = x2 + 2, g(x) = -x, x = 0, and x = 1 .
Area = Top curve – bottom curve
b

A ∫[f(x (x ] x =
=
)−g ) d
a
1

2
(x +2 ( x ] x=
∫[ )− − ) d
0
1

1 1
17
x

x
 + + 2 x = 3 + 2 + 2 = 6
3 2
0
3

2
Find the area of the region bounded by the graphs of
f(x) = 2 – x2 and g(x) = x
First, set f(x) = g(x) to
find their points of
intersection.
2 – x2 = x
0 = x2 + x - 2
0 = (x + 2)(x – 1)
x = -2 and x = 1

∫[
(
1

−
2

) ]

9
2−x −x d fnInt(2 – x – x, x, -2, 1) =
x=
2
2

2
Find the area of the region between the graphs of
f(x) = 3x3 – x2 – 10x and g(x) = -x2 + 2x
Again, set f(x) = g(x)
to find their points of
intersection.

3x3 – x2 – 10x = -x2 + 2x
3x3 – 12x = 0
3x(x2 – 4) = 0
x = 0 , -2 , 2
Note that the two graphs switch at the origin.
Now, set up the two integrals and solve.
0

2

−
2

0

)− ( ) d
g )− ) d
∫[f(x gx] x+∫[ (x f (x] x
0

2

x 1 x x+ −x 1 x x=
∫(3 − 2 )d ∫( 3 + 2 )d
3

−
2

3

0

= 24
x2

A = ∫ [(top curve) − (bottom curve)] dx
x1

y2

A = ∫ [(right curve) − (le f t curve)] dy
y1
1. Find the area of the region bounded by the graphs
of y = 6 + 3 x − x 2 , y = 2 x − 9 , x = -2 , and x = 2.
2.Find the area of the region bounded by the graphs
y = x 2 − 5 , and y = 1 − x .
of
Find the area of the region bounded by the graphs
of x = 3 – y2 and y = x - 1
When you integrate with respect to Y:
•Your functions must all be in terms of y
•Your variable of integration changes to “dy”
•The formula for Area between curves becomes:
b

A = ∫ [(right curve) − (le f t curve)] dy
a
Find the area of the region bounded by the graphs
of x = 3 – y2 and y = x - 1
Area = Right - Left
1

A ∫[ −y )− 1] y
= (
3
(y+ ) d
2

−
2

9
=
2
Find the area of the region bounded by the graphs
of x = y2 and y = x - 2

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0 calc7-1

  • 1. What’s the definite integral used for?
  • 2. f g Area of region between f and g f = = g - Area of region under f(x) b a a g b b (x x ∫[f(x)−g )]d =∫ f ( x)dx f − Area of region under g(x) ∫ g ( x)dx a
  • 3. 7.1 Areas Between Curves To find the area: • divide the area into n strips of equal width • approximate the ith strip by a rectangle with base Δx and height f(xi) – g(xi). • the sum of the rectangle areas is a good approximation • the approximation is getting better as n→∞. y = f(x) y = g(x) The area A of the region bounded by the curves y=f(x), y=g(x), and the lines x=a, x=b, where f and g are continuous and f(x) ≥ g(x) for all x in [a,b], is b A = ∫ [ f ( x) − g ( x)]dx a
  • 4. To find the area between 2 curves (along the x-axis) • Sketch a graph (if you can) • Draw a representative rectangle to determine the upper and lower curves. • Use the formula: Area = Top curve – bottom curve b A ∫[f(x (x ] x = = )−g ) d a
  • 5. Ex. Find the area of the region bounded by the graphs of f(x) = x2 + 2, g(x) = -x, x = 0, and x = 1 . Area = Top curve – bottom curve b A ∫[f(x (x ] x = = )−g ) d a 1 2 (x +2 ( x ] x= ∫[ )− − ) d 0 1 1 1 17 x  x  + + 2 x = 3 + 2 + 2 = 6 3 2 0 3 2
  • 6. Find the area of the region bounded by the graphs of f(x) = 2 – x2 and g(x) = x First, set f(x) = g(x) to find their points of intersection. 2 – x2 = x 0 = x2 + x - 2 0 = (x + 2)(x – 1) x = -2 and x = 1 ∫[ ( 1 − 2 ) ] 9 2−x −x d fnInt(2 – x – x, x, -2, 1) = x= 2 2 2
  • 7. Find the area of the region between the graphs of f(x) = 3x3 – x2 – 10x and g(x) = -x2 + 2x Again, set f(x) = g(x) to find their points of intersection. 3x3 – x2 – 10x = -x2 + 2x 3x3 – 12x = 0 3x(x2 – 4) = 0 x = 0 , -2 , 2 Note that the two graphs switch at the origin.
  • 8. Now, set up the two integrals and solve. 0 2 − 2 0 )− ( ) d g )− ) d ∫[f(x gx] x+∫[ (x f (x] x 0 2 x 1 x x+ −x 1 x x= ∫(3 − 2 )d ∫( 3 + 2 )d 3 − 2 3 0 = 24
  • 9. x2 A = ∫ [(top curve) − (bottom curve)] dx x1 y2 A = ∫ [(right curve) − (le f t curve)] dy y1
  • 10. 1. Find the area of the region bounded by the graphs of y = 6 + 3 x − x 2 , y = 2 x − 9 , x = -2 , and x = 2. 2.Find the area of the region bounded by the graphs y = x 2 − 5 , and y = 1 − x . of
  • 11. Find the area of the region bounded by the graphs of x = 3 – y2 and y = x - 1
  • 12. When you integrate with respect to Y: •Your functions must all be in terms of y •Your variable of integration changes to “dy” •The formula for Area between curves becomes: b A = ∫ [(right curve) − (le f t curve)] dy a
  • 13. Find the area of the region bounded by the graphs of x = 3 – y2 and y = x - 1 Area = Right - Left 1 A ∫[ −y )− 1] y = ( 3 (y+ ) d 2 − 2 9 = 2
  • 14. Find the area of the region bounded by the graphs of x = y2 and y = x - 2