Let ⌊·⌋ denote the greatest integer function. If
∫0e3⌊ex−11⌋dx=α−ln2,
then
α3
is equal to:
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Step-by-Step Solution
Step 1: Simplify the integrand
The given integral involves the greatest integer function of
1/ex−1
. We can simplify the expression inside the greatest integer function by rewriting
1/ex−1
as
e−(x−1)
, which is equal to
e1−x
. This makes the expression easier to work with.
Step 2: Determine the range of
e1−x
To evaluate the greatest integer function, we need to understand the range of values that
e1−x
takes over the integration interval
[0,e3]
. As
x
goes from
0
to
e3
,
1−x
goes from
1
to
1−e3
. Consequently,
e1−x
goes from
e1
down to
e1−e3
. Since
e≈2.718
,
e3≈20.08
, so
1−e3≈−19.08
. Thus,
e1−e3
is a very small positive number, approximately
e−19.08
.
Step 3: Identify integer values of
⌊e1−x⌋
The greatest integer function
⌊e1−x⌋
will take integer values. Since
e1−x
ranges from
e1−e3
(a value between 0 and 1) up to
e
(approximately 2.718), the possible integer values for
⌊e1−x⌋
are
0,1,2
. We need to find the intervals for
x
corresponding to these integer values. For
⌊e1−x⌋=2
, we have
2≤e1−x<3
, which implies
ln2≤1−x<ln3
, leading to
1−ln3<x≤1−ln2
. For
⌊e1−x⌋=1
, we have
1≤e1−x<2
, which implies
0≤1−x<ln2
, leading to
1−ln2<x≤1
. For
⌊e1−x⌋=0
, we have
e1−x<1
, which implies
1−x<0
, leading to
x>1
. Note that
1−ln2≈1−0.693=0.307
and
1−ln3≈1−1.098=−0.098
.
Step 4: Split the integral and evaluate
Based on the intervals determined in the previous step, we can split the integral into three parts. The integral from
0
to
1−ln2
has
⌊e1−x⌋=2
. The integral from
1−ln2
to
1
has
⌊e1−x⌋=1
. The integral from
1
to
e3
has
⌊e1−x⌋=0
. The third integral evaluates to
0
.
Step 5: Calculate the definite integrals
Now we evaluate each definite integral. The first integral is
2x
evaluated from
0
to
1−ln2
, which gives
2(1−ln2)
. The second integral is
x
evaluated from
1−ln2
to
1
, which gives
1−(1−ln2)=ln2
.
Step 6: Sum the results and find
α
Summing the results of the definite integrals, we get
(2−2ln2)+ln2=2−ln2
. We are given that the integral equals
α−ln2
. By comparing, we find that
α=2
.
Step 7: Calculate
α3
Finally, we need to calculate
α3
. Since
α=2
,
α3=23=8
.