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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
67
67
Part of
2005 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 67
Source: created by kunny
7/12/2005
Evaluate
2005
∫
0
1002
d
x
100
2
2
−
x
2
+
100
3
2
−
x
2
+
∫
1002
1003
100
3
2
−
x
2
d
x
∫
0
1
1
−
x
2
d
x
\frac{2005\displaystyle \int_0^{1002}\frac{dx}{\sqrt{1002^2-x^2}+\sqrt{1003^2-x^2}}+\int_{1002}^{1003}\sqrt{1003^2-x^2}dx}{\displaystyle \int_0^1\sqrt{1-x^2}dx}
∫
0
1
1
−
x
2
d
x
2005
∫
0
1002
100
2
2
−
x
2
+
100
3
2
−
x
2
d
x
+
∫
1002
1003
100
3
2
−
x
2
d
x
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