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Today's Calculation Of Integral
2010 Today's Calculation Of Integral
611
611
Part of
2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 611
Source:
6/18/2010
Let
g
(
t
)
g(t)
g
(
t
)
be the minimum value of
f
(
x
)
=
x
2
−
x
f(x)=x2^{-x}
f
(
x
)
=
x
2
−
x
in
t
≤
x
≤
t
+
1
t\leq x\leq t+1
t
≤
x
≤
t
+
1
. Evaluate
∫
0
2
g
(
t
)
d
t
\int_0^2 g(t)dt
∫
0
2
g
(
t
)
d
t
.2010 Kumamoto University entrance exam/Science
calculus
integration
derivative
logarithms
function
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