Do Now: p.381: #8
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Transcript of Do Now: p.381: #8
Do Now: p.381: #8Integrate the two parts separately:
1 22
0 12x dx x dx Shaded Area =
1 23 2
0 1
23 2x x x
1 10 2 4 23 2
56
Integrating with Respect
to ySection 7.2b
Integrating with Respect to ySometimes the boundaries of a region are more easilydescribed by functions of y than by functions of x. In suchcases, we can use horizontal rectangles:
d
c x g y
x f y d
c
x g y
x f y
d
cA f y g y dy
Returning to an example from last class…Find the area of the region R in the first quadrant that is boundedabove by and below by the x-axis and the line .y x 2y x
y x2y x
(4,2)
2 4
1
2
f y g y
y
,g y y
,f y y
0y
Solve our previousequations for x:
2x y2x y 0y
2x y
2x y
2 2
02A y y dy
103
22 3
0
22 3y yy
Returning to an example from last class…And a third way to handle this example: Using geometry formulas.
(4,2)
2 4
1
2
0y
Integrate the square rootfunction from 0 to 4, thensubtract the area of thetriangle on the right:
y x2y x
4
0
1 2 22
A xdx 103
4
3 2
0
2 23x
2
2
Quality Practice ProblemsTry #4 from p.380:
1 2 3 2
012 12 2 2A y y y y dy
43
1 3 2
012 10 2y y y dy
14 3 2
0
1033
y y y 103 13
Quality Practice ProblemsFind the area of the region enclosed by the given functions.
2x y 2x y How about the graph?
(4,2)
(1,–1)
To find the y-coordinates of theIntersection points, solve:
2 2y y 1, 2y y
2 2
12A y y dy
92
8 1 12 4 23 2 3
22 3
1
22 3y yy
Quality Practice ProblemsFind the area of the region enclosed by the given functions.
3y x 2 2x y Use your calculator to graph in [–3, 3] by [–3, 3].
(c,d)
(a,b)
31y x 2 2y x 3 2y x
We could integrate with respect to x,but that would require splitting theregion at x = a.
Instead, let’s integrate from y = b toy = d, handling the entire region atonce…
Quality Practice ProblemsFind the area of the region enclosed by the given functions.
3y x 2 2x y Use your calculator to graph in [–3, 3] by [–3, 3].
(c,d)
(a,b)
31y x 2 2y x 3 2y x
3x y
2 2x y
1 3 2NINT 2 , , 1,DA y y y
1.793003715d
Calculate b and d, storing the valuesin your calculator:
1b
Evaluate the area numerically:
4.215
Do Now: p.381: #8Returning to the “Do Now,” let’s use our new technique!!!
1
02 y y dy Shaded Area =
56
12 3 2
0
1 222 3
y y y
1 22 02 3
Quality Practice Problems23y x Find the area of the region between the curve and
the line by integrating with respect to (a) x, (b) y.1y
2 2
02 3 1A x dx
(2,–1)
Integrate with respect to x:
2 2
02 4 x dx
23
0
12 43
x x
Graph the region:
(–2,–1)
23y x
1y
82 8 03
323
Quality Practice Problems23y x Find the area of the region between the curve and
the line by integrating with respect to (a) x, (b) y.1y
(2,–1)
Graph the region:
(–2,–1) 1y
3
12 3A ydy
Integrate with respect to y:
3
3 2
1
22 33
y
3x y
162 03
323
4y xFind the area of the region in the first quadrant bounded on theleft by the y-axis, below by the line , above left by thecurve , and above right by the curve 2y x
1
01
4xA x dx
Integrate in two parts:
4
1
24x dx
x
Graph the region:
x = 1
1y x
4y x
2y x
1y x
x = 41 42 2
3 2
0 1
2 43 8 8
x xx x x
4y xFind the area of the region in the first quadrant bounded on theleft by the y-axis, below by the line , above left by thecurve , and above right by the curve 2y x
Graph the region:
x = 1
1y x
4y x
2y x
1y x
x = 4
42
1
48xx
123 2
0
23 8
xx x
113
2 1 11 8 2 43 8 8