Area Between Curves: Functions of y
AP Calculus BC · Flashcards · Application
AP CALCULUS BC · FLASHCARDS
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QUESTION
Identify the setup to find the area between
x
=
y
2
x = y^2
x
=
y
2
and
x
=
4
−
y
2
x = 4 - y^2
x
=
4
−
y
2
from
y
=
−
2
y = -2
y
=
−
2
to
y
=
2
y = 2
y
=
2
.
QUESTION
What is the integral setup for
x
=
y
2
−
4
x = y^2 - 4
x
=
y
2
−
4
and
x
=
y
+
2
x = y + 2
x
=
y
+
2
from
y
=
−
3
y = -3
y
=
−
3
to
y
=
3
y = 3
y
=
3
?
QUESTION
State the integral setup for curves
x
=
y
2
+
1
x = y^2 + 1
x
=
y
2
+
1
and
x
=
4
y
−
2
x = 4y - 2
x
=
4
y
−
2
from
y
=
−
1
y = -1
y
=
−
1
to
y
=
3
y = 3
y
=
3
.
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ANSWER
integral from
−
1
to
3
of
[
(
4
y
−
2
)
−
(
y
2
+
1
)
]
dy
\text{integral from } -1 \text{ to } 3 \text{ of } [(4y - 2) - (y^2 + 1)] \text{ dy}
integral from
−
1
to
3
of
[(
4
y
−
2
)
−
(
y
2
+
1
)]
dy
.
(
4
y
−
2
)
(4y - 2)
(
4
y
−
2
)
is rightmost, so subtract
(
y
2
+
1
)
(y^2 + 1)
(
y
2
+
1
)
over the given bounds.
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