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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
27
27
Part of
2005 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 27
Source: 1979 Saitama University
5/28/2005
Let
f
(
x
)
=
t
sin
x
+
(
1
−
t
)
cos
x
(
0
≦
t
≦
1
)
f(x)=t\sin x+(1-t)\cos x\ (0\leqq t\leqq 1)
f
(
x
)
=
t
sin
x
+
(
1
−
t
)
cos
x
(
0
≦
t
≦
1
)
. Find the maximum and minimum value of the following
P
(
t
)
P(t)
P
(
t
)
.
P
(
t
)
=
{
∫
0
π
2
e
x
f
(
x
)
d
x
}
{
∫
0
π
2
e
−
x
f
(
x
)
d
x
}
P(t)=\left\{\int_0^{\frac{\pi}{2}} e^x f(x) dx \right\}\left\{\int_0^{\frac{\pi}{2}} e^{-x} f(x)dx \right\}
P
(
t
)
=
{
∫
0
2
π
e
x
f
(
x
)
d
x
}
{
∫
0
2
π
e
−
x
f
(
x
)
d
x
}
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