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Today's Calculation Of Integral
2006 Today's Calculation Of Integral
131
131
Part of
2006 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 131
Source: Kochi University entrance exam 1966
7/26/2006
For
a
>
0
,
a>0,
a
>
0
,
find the minimum value of
∫
0
1
a
(
a
3
+
4
x
−
a
5
x
2
)
e
a
x
d
x
.
\int_{0}^{\frac{1}{a}}(a^{3}+4x-a^{5}x^{2})e^{ax}\ dx.
∫
0
a
1
(
a
3
+
4
x
−
a
5
x
2
)
e
a
x
d
x
.
calculus
integration
calculus computations