MathDB
Problems
Contests
National and Regional Contests
Japan Contests
Today's Calculation Of Integral
2010 Today's Calculation Of Integral
612
612
Part of
2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 612
Source:
6/28/2010
For
f
(
x
)
=
1
x
(
x
>
0
)
f(x)=\frac{1}{x}\ (x>0)
f
(
x
)
=
x
1
(
x
>
0
)
, prove the following inequality.
f
(
t
+
1
2
)
≤
∫
t
t
+
1
f
(
x
)
d
x
≤
1
6
{
f
(
t
)
+
4
f
(
t
+
1
2
)
+
f
(
t
+
1
)
}
f\left(t+\frac 12 \right)\leq \int_t^{t+1} f(x)\ dx\leq \frac 16\left\{f(t)+4f\left(t+\frac 12\right)+f(t+1)\right\}
f
(
t
+
2
1
)
≤
∫
t
t
+
1
f
(
x
)
d
x
≤
6
1
{
f
(
t
)
+
4
f
(
t
+
2
1
)
+
f
(
t
+
1
)
}
calculus
integration
conics
parabola
quadratics
calculus computations