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Today's Calculation Of Integral
2010 Today's Calculation Of Integral
606
606
Part of
2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 606
Source:
5/4/2010
Find the area of the part bounded by two curves
y
=
x
,
x
+
y
=
1
y=\sqrt{x},\ \sqrt{x}+\sqrt{y}=1
y
=
x
,
x
+
y
=
1
and the
x
x
x
-axis.1956 Tokyo Institute of Technology entrance exam
calculus
integration
geometry
calculus computations