Do Now: p.381: #8Integrate the two parts separately:
1 22
0 12x dx x dx Shaded Area =
1 23 2
0 1
23 2
x xx
1 10 2 4 2
3 2
5
6
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 yd
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
10
322 3
0
22 3
y 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
12 2
2A xdx
10
3
43 2
0
22
3x
2
2
Quality Practice Problems
Try #4 from p.380:
1 2 3 2
012 12 2 2A y y y y dy
4
3
1 3 2
012 10 2y y y dy
14 3 2
0
103
3y y y 10
3 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
9
2
8 1 12 4 2
3 2 3
22 3
1
22 3
y 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 =
5
6
12 3 2
0
1 22
2 3y 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 4
3x x
Graph the region:
(–2,–1)
23y x
1y
82 8 0
3
32
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,–1)
Graph the region:
(–2,–1) 1y
3
12 3A ydy
Integrate with respect to y:
3
3 2
1
22 3
3y
3x y
162 0
3
32
3
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
4
xA x dx
Integrate in two parts:
4
1
2
4
xdx
x
Graph the region:
x = 1
1y x
4y x
2y x
1y x
x = 41 42 2
3 2
0 1
24
3 8 8
x xx x x