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Today's Calculation Of Integral
2006 Today's Calculation Of Integral
118
118
Part of
2006 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 118
Source: Meiji University entrance exam 1978
7/7/2006
Let
f
(
x
)
f(x)
f
(
x
)
be the function defined for
x
≥
0
x\geq 0
x
≥
0
which satisfies the following conditions. (a)
f
(
x
)
=
{
x
(
0
≤
x
<
1
)
2
−
x
(
1
≤
x
<
2
)
f(x)=\begin{cases}x \ \ \ \ \ \ \ \ ( 0\leq x<1) \\ 2-x \ \ \ (1\leq x <2) \end{cases}
f
(
x
)
=
{
x
(
0
≤
x
<
1
)
2
−
x
(
1
≤
x
<
2
)
(b)
f
(
x
+
2
n
)
=
f
(
x
)
(
n
=
1
,
2
,
⋯
)
f(x+2n)=f(x) \ (n=1,2,\cdots)
f
(
x
+
2
n
)
=
f
(
x
)
(
n
=
1
,
2
,
⋯
)
Find
lim
n
→
∞
∫
0
2
n
f
(
x
)
e
−
x
d
x
.
\lim_{n\to\infty}\int_{0}^{2n}f(x)e^{-x}\ dx.
lim
n
→
∞
∫
0
2
n
f
(
x
)
e
−
x
d
x
.
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