Let R be the region in the first quadrant bounded above by the graph of y=(6x+4)12 the line y=2x and the y axis, how do you find the area of region R?
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Answer:
The area is
=
20
9
u
2
Explanation:
The point of intersection of the line
y
=
2
x
with the curve
y
=
√
6
x
+
4
is
√
6
x
+
4
=
2
x
6
x
+
4
=
4
x
2
2
x
2
−
3
x
−
2
=
0
x
=
3
±
√
9
+
16
4
=
3
±
5
4
x
=
2
or
x
=
−
1
2
∫
√
6
x
+
4
=
(
6
x
+
4
)
3
2
3
2
⋅
6
=
1
9
(
6
x
+
4
)
3
2
+
C
∫
2
x
d
x
=
2
⋅
x
2
2
+
C
Therefore,
The area is
A
=
∫
2
0
(
√
6
x
+
4
−
2
x
)
d
x
=
[
1
9
(
6
x
+
4
)
3
2
−
x
2
]
2
0
=
(
1
9
(
16
)
3
2
−
4
)
−
(
1
9
⋅
4
3
2
)
=
28
9
−
8
9
=
20
9
u
2
graph{(y-sqrt(6x+4))(y-2x)=0 [-3.9, 10.146, -0.91, 6.11]}
Step-by-step explanation:
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