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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
696
696
Part of
2011 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 696
Source:
5/28/2011
Let
P
(
x
)
,
Q
(
x
)
P(x),\ Q(x)
P
(
x
)
,
Q
(
x
)
be polynomials such that :
∫
0
2
{
P
(
x
)
}
2
d
x
=
14
,
∫
0
2
P
(
x
)
d
x
=
4
,
∫
0
2
{
Q
(
x
)
}
2
d
x
=
26
,
∫
0
2
Q
(
x
)
d
x
=
2.
\int_0^2 \{P(x)\}^2dx=14,\ \int_0^2 P(x)dx=4,\ \int_0^2 \{Q(x)\}^2dx=26,\ \int_0^2 Q(x)dx=2.
∫
0
2
{
P
(
x
)
}
2
d
x
=
14
,
∫
0
2
P
(
x
)
d
x
=
4
,
∫
0
2
{
Q
(
x
)
}
2
d
x
=
26
,
∫
0
2
Q
(
x
)
d
x
=
2.
Find the maximum and the minimum value of
∫
0
2
P
(
x
)
Q
(
x
)
d
x
\int_0^2 P(x)Q(x)dx
∫
0
2
P
(
x
)
Q
(
x
)
d
x
.
calculus
integration
algebra
polynomial
inequalities
function
linear algebra