4. Find the area of the region enclosed by y = -z² +6, y = 1, y = 5x, and r2 0.

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Just need my work checked, am I placing the functions in the right spot to begin with? How do I know what’s f1(x),f2(x) etc.
4. Find the area of the region enclosed by y = -x² +6, y = x, y = 5x, and r2 0.
Y= - x+6
Y = SX
Intersect
Points
(f,6) - f2(x) dx
dx
ーx?+o -5×
ーx?+6-SX =0
-x2 5x tlo =o
Scxルー)4
x²+ Sx -6=D0
(マー1)(x +6 =o
2.
S (x-5x) dx
(-x²-x+6) dx
Simplify
(x2+Xー6)dx
に(4x) dx
- x2+6=X
- x2_x +Lo=o
2.
り
dx
(ベ-3(x+3)
Substitute in f (x)
- le
5X=X
3.
34
Sx -x =0
し
4x=0
ニ
36
G) +
H - キ
-8
31
+ 36
こ67
Transcribed Image Text:4. Find the area of the region enclosed by y = -x² +6, y = x, y = 5x, and r2 0. Y= - x+6 Y = SX Intersect Points (f,6) - f2(x) dx dx ーx?+o -5× ーx?+6-SX =0 -x2 5x tlo =o Scxルー)4 x²+ Sx -6=D0 (マー1)(x +6 =o 2. S (x-5x) dx (-x²-x+6) dx Simplify (x2+Xー6)dx に(4x) dx - x2+6=X - x2_x +Lo=o 2. り dx (ベ-3(x+3) Substitute in f (x) - le 5X=X 3. 34 Sx -x =0 し 4x=0 ニ 36 G) + H - キ -8 31 + 36 こ67
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