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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
41
41
Part of
2005 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 41
Source: 1988 Wakayama Medical University
6/10/2005
Evaluate
∫
0
a
2
a
x
−
x
2
d
x
(
a
>
0
)
\int_0^a \sqrt{2ax-x^2}\ dx \ (a>0)
∫
0
a
2
a
x
−
x
2
d
x
(
a
>
0
)
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