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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
704
704
Part of
2011 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 704
Source: 2011 Fukui University entrance exam/Medicine
6/3/2011
A function
f
n
(
x
)
(
n
=
0
,
1
,
2
,
3
,
⋯
)
f_n(x)\ (n=0,\ 1,\ 2,\ 3,\ \cdots)
f
n
(
x
)
(
n
=
0
,
1
,
2
,
3
,
⋯
)
satisfies the following conditions:(i)
f
0
(
x
)
=
e
2
x
+
1
f_0(x)=e^{2x}+1
f
0
(
x
)
=
e
2
x
+
1
.(ii)
f
n
(
x
)
=
∫
0
x
(
n
+
2
t
)
f
n
−
1
(
t
)
d
t
−
2
x
n
+
1
n
+
1
(
n
=
1
,
2
,
3
,
⋯
)
.
f_n(x)=\int_0^x (n+2t)f_{n-1}(t)dt-\frac{2x^{n+1}}{n+1}\ (n=1,\ 2,\ 3,\ \cdots).
f
n
(
x
)
=
∫
0
x
(
n
+
2
t
)
f
n
−
1
(
t
)
d
t
−
n
+
1
2
x
n
+
1
(
n
=
1
,
2
,
3
,
⋯
)
.
Find
∑
n
=
1
∞
f
n
′
(
1
2
)
.
\sum_{n=1}^{\infty} f_n'\left(\frac 12\right).
∑
n
=
1
∞
f
n
′
(
2
1
)
.
calculus
integration
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