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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
58
Today's Calculation of Integral 58
Today's Calculation of Integral 58
Source: 1998 Tohoku University
June 30, 2005
calculus
integration
calculus computations
Problem Statement
Let
f
(
x
)
=
e
x
e
x
+
1
f(x)=\frac{e^x}{e^x+1}
f
(
x
)
=
e
x
+
1
e
x
Prove the following equation.
∫
a
b
f
(
x
)
d
x
+
∫
f
(
a
)
f
(
b
)
f
−
1
(
x
)
d
x
=
b
f
(
b
)
−
a
f
(
a
)
\int_a^b f(x)dx+\int_{f(a)}^{f(b)} f^{-1}(x)dx=bf(b)-af(a)
∫
a
b
f
(
x
)
d
x
+
∫
f
(
a
)
f
(
b
)
f
−
1
(
x
)
d
x
=
b
f
(
b
)
−
a
f
(
a
)
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