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National and Regional Contests
Japan Contests
Today's Calculation Of Integral
2010 Today's Calculation Of Integral
623
623
Part of
2010 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 623
Source:
7/7/2010
Find the continuous function satisfying the following equation.
∫
0
x
f
(
t
)
d
t
+
∫
0
x
t
f
(
x
−
t
)
d
t
=
e
−
x
−
1.
\int_0^x f(t)dt+\int_0^x tf(x-t)dt=e^{-x}-1.
∫
0
x
f
(
t
)
d
t
+
∫
0
x
t
f
(
x
−
t
)
d
t
=
e
−
x
−
1.
1978 Shibaura Institute of Technology entrance exam
calculus
integration
function
calculus computations