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Today's Calculation Of Integral
2010 Today's Calculation Of Integral
618
Today's calculation of Integral 618
Today's calculation of Integral 618
Source:
July 1, 2010
calculus
integration
trigonometry
calculus computations
Problem Statement
Find the minimu value of
1
π
∫
−
π
2
π
2
{
x
cos
t
+
(
1
−
x
)
sin
t
}
2
d
t
.
\frac{1}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \{x\cos t+(1-x)\sin t\}^2dt.
π
1
∫
−
2
π
2
π
{
x
cos
t
+
(
1
−
x
)
sin
t
}
2
d
t
.
2010 Ibaraki University entrance exam/Science
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