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Today's Calculation Of Integral
2012 Today's Calculation Of Integral
809
809
Part of
2012 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 809
Source: 2012 Kobe University entrance exam/Humanities
3/16/2012
For
a
>
0
a>0
a
>
0
, denote by
S
(
a
)
S(a)
S
(
a
)
the area of the part bounded by the parabolas
y
=
1
2
x
2
−
3
a
y=\frac 12x^2-3a
y
=
2
1
x
2
−
3
a
and
y
=
−
1
2
x
2
+
2
a
x
−
a
3
−
a
2
y=-\frac 12x^2+2ax-a^3-a^2
y
=
−
2
1
x
2
+
2
a
x
−
a
3
−
a
2
. Find the maximum area of
S
(
a
)
S(a)
S
(
a
)
.
calculus
integration
geometry
conics
parabola
calculus computations