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Today's Calculation Of Integral
2006 Today's Calculation Of Integral
162
162
Part of
2006 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 162
Source: created by kunny
10/21/2006
Let
f
(
x
)
f(x)
f
(
x
)
be the function such that
f
(
x
)
>
0
f(x)>0
f
(
x
)
>
0
at
x
≥
0
x\geq 0
x
≥
0
and
{
f
(
x
)
}
2006
=
∫
0
x
f
(
t
)
d
t
+
1.
\{f(x)\}^{2006}=\int_{0}^{x}f(t) dt+1.
{
f
(
x
)
}
2006
=
∫
0
x
f
(
t
)
d
t
+
1.
Find the value of
{
f
(
2006
)
}
2005
.
\{f(2006)\}^{2005}.
{
f
(
2006
)
}
2005
.
calculus
integration
function
calculus computations