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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
49
Today's Calculation of Integral 49
Today's Calculation of Integral 49
Source: created by kunny
June 19, 2005
calculus
integration
trigonometry
geometry
function
calculus computations
Problem Statement
For
x
≥
0
x\geq 0
x
≥
0
, Prove that
∫
0
x
(
t
−
t
2
)
sin
2002
t
d
t
<
1
2004
⋅
2005
\int_0^x (t-t^2)\sin ^{2002} t \,dt<\frac{1}{2004\cdot 2005}
∫
0
x
(
t
−
t
2
)
sin
2002
t
d
t
<
2004
⋅
2005
1
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