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Today's Calculation Of Integral
2010 Today's Calculation Of Integral
608
608
Part of
2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 608
Source:
5/30/2010
For
a
>
0
a>0
a
>
0
, find the minimum value of
∫
0
1
a
x
2
+
(
a
2
+
2
a
)
x
+
2
a
2
−
2
a
+
4
(
x
+
a
)
(
x
+
2
)
d
x
.
\int_0^1 \frac{ax^2+(a^2+2a)x+2a^2-2a+4}{(x+a)(x+2)}dx.
∫
0
1
(
x
+
a
)
(
x
+
2
)
a
x
2
+
(
a
2
+
2
a
)
x
+
2
a
2
−
2
a
+
4
d
x
.
2010 Gakusyuin University entrance exam/Science
calculus
integration
calculus computations