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Today's Calculation Of Integral
2006 Today's Calculation Of Integral
145
145
Part of
2006 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 145
Source: Yamagata University entrance exam 1980
8/28/2006
Find the minimum value of
∫
x
x
+
l
(
t
+
1
t
)
d
t
(
x
>
0
,
l
>
0
)
.
\int_{x}^{x+l}\left(t+\frac{1}{t}\right)dt \ (x>0,\ l>0).
∫
x
x
+
l
(
t
+
t
1
)
d
t
(
x
>
0
,
l
>
0
)
.
calculus
integration
logarithms
absolute value
calculus computations