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Today's Calculation Of Integral
2010 Today's Calculation Of Integral
633
633
Part of
2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 633
Source:
8/4/2010
Let
f
(
x
)
f(x)
f
(
x
)
be a differentiable function. Find the value of
x
x
x
for which
{
f
(
x
)
}
2
+
(
e
+
1
)
f
(
x
)
+
1
+
e
2
−
2
∫
0
x
f
(
t
)
d
t
−
2
f
(
x
)
∫
0
x
f
(
t
)
d
t
+
2
{
∫
0
x
f
(
t
)
d
t
}
2
\{f(x)\}^2+(e+1)f(x)+1+e^2-2\int_0^x f(t)dt-2f(x)\int_0^x f(t)dt+2\left\{\int_0^x f(t)dt\right\}^2
{
f
(
x
)
}
2
+
(
e
+
1
)
f
(
x
)
+
1
+
e
2
−
2
∫
0
x
f
(
t
)
d
t
−
2
f
(
x
)
∫
0
x
f
(
t
)
d
t
+
2
{
∫
0
x
f
(
t
)
d
t
}
2
is minimized.1978 Tokyo Medical College entrance exam
calculus
integration
calculus computations