# 1. Evaluate the integral = ( ```Calculus Challenge 2001
R
1. Evaluate the integral 1 dxx3
The method of partial fractions and completing
the square yields
u? 32
x
?
x3 = ( x ? x2 ?x ) = ( x ? u2 34 ) where u = x ? . This leads
to
R 3dx = (ln(x + 1) ? ln(u + ) + p tan? ( pu )) = (ln( p x ) +
3
x2 ?x
2
px ? xp?
p
p this ?we p?obtain p R 13tandx3 (= ( p). From
? 3tan ( )) = ( + ) = .
X
0
1
+1
1
3
1
+1
1 2
0
2
+1
1
3
+1
+1
1+
1
3
1
2
1
+1
2
1
2
+
3
4
3 2
2 3
1
3
3
3
+1
1
3
1
+1
1
3
1
3
3
2
1
1
2
1
6
3
2
9
2. Show that xe ex for all positive real numbers x and determine all
number x for which equality holds.
1
1
1
The
inequality
is
equivalent
to
x x e e . Let f (x) = x x . Then f 0 (x) =
1
x x ( ?xln2 x ). This is positive if 0 &lt; x &lt; e and is negative if e &lt; x.
Therefore, f (x) has an absolute maximum at x = e and so f (x) f (e)
for all x &gt; 0 and equality holds if, and only if, x = e. Thus xe ex for
all x &gt; 0 and equality holds if, and only if, x = e.
1
( )
3. Let &gt; ?1 and &gt; ?1. Calculate limn!1 n?R nn
Pn k 1 dx limn!1 n? nn = limn!1 Pkn=1 nnk = R01 xx dx
=
1 +2 +
1 +2 +
1 +2 +
1 +2 +
+
+
k=1
(
)
(
)
0
+
+
+1
+1
ln
4. Find the sum of the following series: P1
n
en
n
P nxn =
P1 ln n = ln(2) P1 ne?n = ln(2)f ( ) where
f (x) = 1
n
n
nP e
e
n )0 = x(
0 = x 2 for jxj &lt; 1. Thus the value of the series
x( 1
x
)
n
?x
?x
e?1 ln
eln
is ?e?1 2 = e? 2 .
(2n )
=1
1
(2 )
=1
1
=0
(1
(2)
)
(
1
(2)
1)
=1
(1
)
=1
R
5. Find the antiderivative: pedx
a ?ex
x
The substitution sin = ee a22 reduces the integral to
R a cosd = 2e ?2a R cscd = ?2e ?2a lnjcot+cscj+C = ?2e ?2a ln( pea?exx
e 2 sincos
e2
p
?a
?a
a
C = xe 2 ? 2e 2 ln(e 2 + ea ? ex ) + C
2
6. Does the series P1
n ne
This converges because of the integral test. Using the substitution
with integration by parts,
Ru1=xex?ptogether
x dx = R 1 2u e?u du = 2(?u e?u ? 3u e?u ? 6ue?u ? 6e?u )j1 =
p
=1
2
1
32
e
1
3
3
2
1
e
a
+ 2
)+
```