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Today's Calculation Of Integral
2010 Today's Calculation Of Integral
625
Today's calculation of Integral 625
Today's calculation of Integral 625
Source:
July 8, 2010
calculus
integration
limit
trigonometry
calculus computations
Problem Statement
Find
lim
t
→
0
1
t
3
∫
0
t
2
e
−
x
sin
x
t
d
x
(
t
≠
0
)
.
\lim_{t\rightarrow 0}\frac{1}{t^3}\int_0^{t^2} e^{-x}\sin \frac{x}{t}\ dx\ (t\neq 0).
lim
t
→
0
t
3
1
∫
0
t
2
e
−
x
sin
t
x
d
x
(
t
=
0
)
.
2010 Kumamoto University entrance exam/Medicine
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