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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
72
72
Part of
2005 Today's Calculation Of Integral
Problems
(1)
Today's Calculation of Integral 72
Source: 1992 Chiba University
7/16/2005
Let
f
(
x
)
f(x)
f
(
x
)
be a continuous function satisfying
f
(
x
)
=
1
+
k
∫
−
π
2
π
2
f
(
t
)
sin
(
x
−
t
)
d
t
(
k
:
c
o
n
s
t
a
n
t
n
u
m
b
e
r
)
f(x)=1+k\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(t)\sin (x-t)dt\ (k:constant\ number)
f
(
x
)
=
1
+
k
∫
−
2
π
2
π
f
(
t
)
sin
(
x
−
t
)
d
t
(
k
:
co
n
s
t
an
t
n
u
mb
er
)
Find the value of
k
k
k
for which
∫
0
π
f
(
x
)
d
x
\int_0^{\pi} f(x)dx
∫
0
π
f
(
x
)
d
x
is maximized.
calculus
integration
function
trigonometry
calculus computations