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Today's Calculation Of Integral
2012 Today's Calculation Of Integral
822
822
Part of
2012 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 822
Source: 2012 Iwate University entrance exam/Engineering
6/13/2012
For
n
=
0
,
1
,
2
,
⋯
n=0,\ 1,\ 2,\ \cdots
n
=
0
,
1
,
2
,
⋯
, let
a
n
=
∫
n
n
+
1
{
x
e
−
x
−
(
n
+
1
)
e
−
n
−
1
(
x
−
n
)
}
d
x
,
a_n=\int_{n}^{n+1} \{xe^{-x}-(n+1)e^{-n-1}(x-n)\}\ dx,
a
n
=
∫
n
n
+
1
{
x
e
−
x
−
(
n
+
1
)
e
−
n
−
1
(
x
−
n
)}
d
x
,
b
n
=
∫
n
n
+
1
{
x
e
−
x
−
(
n
+
1
)
e
−
n
−
1
}
d
x
.
b_n=\int_{n}^{n+1} \{xe^{-x}-(n+1)e^{-n-1}\}\ dx.
b
n
=
∫
n
n
+
1
{
x
e
−
x
−
(
n
+
1
)
e
−
n
−
1
}
d
x
.
Find
lim
n
→
∞
∑
k
=
0
n
(
a
k
−
b
k
)
.
\lim_{n\to\infty} \sum_{k=0}^n (a_k-b_k).
lim
n
→
∞
∑
k
=
0
n
(
a
k
−
b
k
)
.
calculus
integration
limit
calculus computations