Ruben Costa Santos
Section 10.1
3.
Is the folowing integral improper?
The integral is improper because the integrand function is unbound when
,
10.
Evaluate the improper integral:
Let us start by finding the antiderivative. Using the substitution
and
du=dx/x we get:
11.
Evaluate the improper integral
then:
Notice that
Since this limit exist the improper integral is convergente:
14.
Evaluate the improper integral
The antiderivative can be found be integration by parts in a similar way to example 8, page 497 of the book:
Now we integrate between
and t:
and take the limit
29.
Evaluate the improper integral
The antiderivative is easy
then integrate between a small
and 8 and take the limit
and we see that the integral converges to 6.
31.
Evalute the improper integral
Let us start by finding the antiderivative:
This time the integrand is unbound for
so we are interested in the limit
37. Evaluate the improper integral
Let us start by fiding the antiderivative:
were we made the sustitution u=ex and
.
The integral can be breaked into two parts and we look at them separately:
Let us look at the second part. We want to look at the limit
The limit for the other integral as exactly the same value, as you can see by
doing it in the same way than above. This is not a surprise since the integrand is
even in x (this means that
f(x)=f(-x)) and we can relate the two integrals by
a change of variavel u=-x and du=-dx :
Then we conclude that the improper integral is convergent to the value
41. a)
Show that the improper integral
diverges.
The antiderivative is just:
b)
Show that the
with p>1 converges.
Let us look, as usual, to the integral between 1 and a certain t:
Since this the limit is finite the integral converges.
c)
Show that the
with p<1 diverges.
The antiderivative is the same than above but now we have:
44.a)
Suppose that
were c is a real number. Then the area from
x=0 to x=t between the
graph of f(x) and the x axis is larger
then the area from 0 to t between the line y=c and the x axis. In terms of
integrals:
b)
Now suppose that
.
Then the area from
x=1 to x=t between the
graph of g(x) and the x axis is larger
then the area from 1 to t between the graph of y=x-2 and the x axis.
In terms of integrals:
We know from exercise 41.b) that the integral
converges then the limit of the integral for
must verify
c)
Suppose that
for
.
Then we must have, following the same
reasoning on the areas, that