We want to find the area under the curve
y equals 1.5 x to the power of negative 2.5,
graphed here in blue, from x equals one to x equals t,
and evaluate for t equals two and t equals five.
Then determine the total area under the curve
for x greater than or equal to positive one.
So we'll have three separate integrals here.
For the first integral, we'll integrate from x equals one
to x equals two.
So we'll have the integral of, let's write 1.5
as three halves, x to the power of, let's write negative
2.5 as negative five halves.
We'll integrate respect to x, make it first from one to two,
which will give us the area of the shaded region here.
So now we'll find the antiderivative of three halves x
to the power of negative five halves.
So we have three halves times x to the power
of negative five halves plus one, that would be
negative three halves divided by negative three halves.
First we'll simplify this, divided by negative
three halves is equivalent to multiplying by the
reciprocal of negative two thirds.
So we have three halves times negative two thirds,
and let's write x to the negative three halves as
one over x to the positive three halves.
So now we'll simplify the three, simplify the two simplify.
So we end up with negative one divided by x to the power of
three halves.
So now notice when x is two, we have negative one
divided by two to the three halves, minus, when x is one,
we have negative one divided by one to the three halves.
So for the exact value, notice how one of the three halves
is one, so we have minus negative one becomes positive one.
So the exact value would be one minus one divided by
two to the three halves.
Let's also get our decimal approximation though.
We have one minus one divided by two raised to the power of
positive three halves, enter.
So the approximate value would be 0.6464,
which you know would be the approximate area
of the shaded region.
Now for the next two integrals, we'll save quite a bit
of time, because we know the antiderivative
of three halves x to the power of negative five halves
is negative one divided by x to the three halves.
So for the second part, we'll integrate from x equals one
to x equals five.
So, again, we'll have the integral of three halves
x to the power of negative five halves, integrated with
respect to x from one to five now.
So we know the antiderivative is negative one divided by
x to the power of three halves.
So when x is five, we have negative one divided by
five to the three halves, minus when x is one, we have
negative one divided by one to the three halves.
So, this ends up being plus one, so we have one minus
one divided by five to the three halves.
This would be the exact value.
Let's go ahead and get the decimal approximation.
We can go ahead and just press second enter,
which brings up the last expression, and we can just
change this two to a five.
So the approximate value would be 0.9106,
which would be the approximate area of the shaded region
here, on the closed interval from one to five.
And now, at the last integral, which will be
an improper integral, we want to integrate
from one to positive infinity.
So we'd have the integral, again, of three halves,
x to the power of negative five halves, with respect to x
from one to infinity.
So we have an improper integral here, so go ahead and
rewrite this as the limit as b approaches infinity
of the integral from one to b of three halves,
x to the power of negative five halves, integrate with
respect to x.
So because we had an improper integral, we would replace
the upper limit of integration, here, with b, and then we
took the limit as b approach infinity throughout
this integral here.
And if this limit exists, then we say this improper integral
converges, and if it doesn't exist, we say it diverges.
So now we have the limit as b approaches infinity of,
we know the antiderivative is going to be negative one
divided by x to the three halves, now limits of integration
are from one to b, so when x is equal to b we have negative
one divided by b to the three halves.
And then when x is positive one, we have minus negative one
divided by one to the three halves.
So notice here, as b approaches infinity, we have this
fraction that has negative one in the numerator,
and the denominator increases without bound,
so this fraction approaches zero, and this fraction is not
affected by b, so we end up having minus negative one,
or just positive one.
And since this limit is equal to positive one,
this improper integral converges, and we can say that the
integral from one to infinity of three halves x to the power
of negative five halves, integrative respect to x does equal
one, meaning the area under this curve for x greater than
or equal to one, is equal to positive one.
I hope you found this helpful.