MathDB
Today's calculation of Integral 67

Source: created by kunny

July 12, 2005
calculusintegrationgeometrycalculus computations

Problem Statement

Evaluate 2005∫01002dx10022−x2+10032−x2+∫1002100310032−x2dx∫011−x2dx\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}
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