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Today's Calculation Of Integral
2011 Today's Calculation Of Integral
688
688
Part of
2011 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 688
Source:
3/1/2011
For a real number
x
x
x
, let
f
(
x
)
=
∫
0
π
2
∣
cos
t
−
x
sin
2
t
∣
d
t
f(x)=\int_0^{\frac{\pi}{2}} |\cos t-x\sin 2t|\ dt
f
(
x
)
=
∫
0
2
π
∣
cos
t
−
x
sin
2
t
∣
d
t
.(1) Find the minimum value of
f
(
x
)
f(x)
f
(
x
)
.(2) Evaluate
∫
0
1
f
(
x
)
d
x
\int_0^1 f(x)\ dx
∫
0
1
f
(
x
)
d
x
.2011 Tokyo Institute of Technology entrance exam, Problem 2
calculus
integration
trigonometry
logarithms
calculus computations