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Drill
• Find the area between the x-axis and the
graph of the function over the given interval:
• y = sinx over [0, π]
 sin xdx

0

 cos x
0
 cos   cos 0  2
• y=
4x-x3
over [0,3]
2
3
 (4 x  x
) dx 
3
0
2
2
2x 
2
x
4
4
0
 (4 x  x
3
 2x 
2
2
4   6 . 25  10 . 25
x
4
4
3
) dx
7.2: Applications of Definite
Integrals
Day #1 Homework: page 395-396 (1-10)
Day #2 Homework: Page 396-397, 11-14, 2737 (odd)
What you’ll learn about…
•
•
•
•
•
Area between curves
Ares enclosed by intersecting curves
Boundaries with changing functions
Integrating with respect to y
Saving time with geometric formulas.
Area Between Curves
• If f and g are continuous with f(x) > g(x)
through [a,b], then the area between the
curves y = f(x) and y = g(x) from a to b is the
integral of [f-g] from a to b,
b
A 
 [ f ( x )  g ( x )] dx
a
Applying the Definition
• Find the area of the region between the and y = sec2 x and y = sinx from x = 0
to x = π/4.
• When graphed, you can see that y = sec2 x is above y = sinx on [0, π/4].
 /4
 (sec
2
 /4
x  sin x ) dx
0
tan x  cos x
0
(tan

4
 cos

) 1
4
1
2
(tan 0  cos 0 )  0  1  1
2
2
2
1 
2
2
units
2
•
•
•
Area of an Enclosed Region
When a region is enclosed by intersecting curves, the intersection points give the
limits of integration.
Find the area of the region enclosed by the parabola y = 2 – x2 and the line y = -x.
Solution: graph the curves to determine the x-values of the intersections and if the
parabola or the line is on top.
2
 [( 2  x )  (  x )] dx
2
1
2
 ( 2  x  x ) dx
2
1
2
2x 
1
x
3
3

x
2
2
( 4  8 / 3  2 )  (  2  1 / 3  1 / 2 )  5 . 5 units
2
Using a Calculator
• Find the area of the region enclosed by the graphs of
y = 2cosx and y = x2-1.
• Solution: Graph to determine the intersections. Don’t
forget to use Zoom 7: Trig (radian mode, please)
 X = ± 1.265423706
 fnInt(2cosx –(x2-1), x, -1.265423706, 1.265423706)
 4.994 units2
Boundaries with Changing Functions
• If a boundary of a region is defined by more than one function, we can
partition the region into subregions that correspond to the function
changes.
• Find the area of the region R in the first quadrant bounded by y = (x)1/2
and below by the x-axis AND the line y = x -2
2
Region A:
2 2
 (x
1/ 2
) dx
0
x
 1 . 8856
3/2
0 3
4
Region B:
 (x
1/ 2
 ( x  2 )) dx
2
4 2
2 3
x
3/2

1
x  2x
2
2
 1 . 4447733
Drill: Find the x and y coordinates of all points where
the graphs of the given functions intersect
• y = x2 – 4x and x + 6
• y = ex and y = x + 1
• y = x2 – πx and y = sin x
• (-1, 5) and (6, 12)
• (0,1)
• (0,0)and (3.14, 0)
Integrating with Respect to y
• Sometimes the boundaries of a region are
more easily described by functions of y than
by functions of x. We can use approximating
rectangles that are horizontal rather than
vertical and the resulting basic formula has y
in place of x.
Integrating with Respect to y
• Find the area of the region R in the first
quadrant bounded by y = (x)1/2 and below by
the x-axis AND the line y = x -2 by integrating
with respect to x.
• y = x – 2 becomes y + 2 = x and y = (x)1/2
becomes y2 = x, y > 0
2
 ( y  2  y ) dy
2
0
2
y
0
(2)
2
2
 2y 
2
2
 2(2) 
y
3
3
(2)
3
3
3
 (0) 2
(0) 

 
 2(0) 
3 
 2
Making the Choice
• Find the area of the region enclosed
by the graph
3
x = y2 – 2 (To graph…. y 1  x
y2  
x2
y = x3 and
• It makes more sense to integrate with respect to y because
you will not have to split the region.
• The lower limit is when the y – value = -1
• The upper limit needs
to be found in the calc.
y1  x
y2 
y = 1.79
3
x2
Solution
• Solve both equations for x.
 y1/3 = x and x = y2 -2
• Set up your definite integral: right – left
1 . 79
 (y
1/3
 ( y  2 )) dy
2
0
• Evaluate using the calculator.
 fnInt(x1/3 – x2 + 2, x, -1, 1.79)
 4.21 units2
Using Geometry
• Another way to complete find the area of the region R in the
first quadrant bounded by y = (x)1/2 and below by the x-axis AND
the line y = x -2 by integrating with respect to x.
• You can find the area under y = x to the x-axis from 0 to 4 and
then subtract the area of the triangle formed by the line y = x -2
and the x-axis. 4
4
 (x
4
3
0
x
3/2
x

2
) dx 
0
4
2
1/ 2
2
2
2
 2x
 ( x  2 ) dx
(
2
3
4
3/2
2
 42

2
)  (
 2  4)  (
 2  2)
2
 2


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