MathDB
Problems
Contests
National and Regional Contests
Japan Contests
Today's Calculation Of Integral
2010 Today's Calculation Of Integral
612
Today's calculation of Integral 612
Today's calculation of Integral 612
Source:
June 28, 2010
calculus
integration
conics
parabola
quadratics
calculus computations
Problem Statement
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
)
}
Back to Problems
View on AoPS