We want to evaluate the integral
and then determine if the integral
is convergent or divergent.
The given integral is an improper integral
because the upper limit of integration is unbounded
or approaching infinity.
We begin by writing the improper integral as a limit.
The given improper integral is equal to the limit
as b approaches, infinity of the integral from 25
to b of e raise the power of negative square root x.
Let's write that as negative x to the power of 1/2
divided by the square root of x
which we'll write as x the power of 1/2
and then we have dx.
Notice how we replace the upper limit of integration with b
and then we're taking the limit as b approaches infinity
of the integral from 25 to b.
From here, to evaluate the integral,
we will have to perform u substitution.
We will let u equal the exponent of negative x
the power of 1/2 and therefore du is equal
to the derivative of negative x the power of 1/2
times dx which is negative 1/2 times x to the power of 1/2
minus one, which is negative 1/2,
times dx giving us du equals,
let's write this as negative 1/2 times one divided by x
the power of positive 1/2 times dx.
And now let's go back and look at the integral.
We can replace negative x to the power of 1/2 with u
and we're left with one divided by x to the power of 1/2 dx.
So let's go ahead and solve this last equation
for one divided by x the power of 1/2 dx
by multiplying both sides by negative two
which gives us negative two du equals one divided by x
the power of 1/2 times dx.
Notice negative two times negative 1/2 is positive one.
So now we can replace one divided by x
the power of 1/2 dx with negative two du.
So this gives us the limit as b approaches infinity
of the integral.
Now we need to be careful here
with the limits of integration from 25 to b,
these are x values, not u values.
So I'm going to write the lower limit as x equals 25
and the upper limit as x equals b.
Just to emphasize, these are x values not u values.
And again, now we know one divided by x
the power of 1/2 dx is equal to negative du.
So I'm going to go ahead and factor out the negative two.
Then we have du and then since u is equal to negative x
to the power of 1/2, the integral function
with respect to u is just either of the u.
This gives us negative two times the limit
as b approaches infinity of well, the antiderivative
of e of the u is e of the u, which with respect to x
would be e the power of negative x to the power of 1/2.
And again, the limits integration for x are from 25 to b.
For the next step we determine big f of b minus big f of 25.
Well, big f of b is e raised to the power of negative b
to the power of 1/2 minus big f of 25
is e to the power of negative 25 to the power of 1/2.
For the next step let's rewrite this
using positive exponents
and also write the exponents of 1/2 back as square roots.
So we have negative two times the limit
as b approaches infinity of,
e to the power of negative b to the power of 1/2
is equal to one divided by e
raise to the power of square root b.
And then we have minus e to the power
of negative 25 to the power of 1/2 is equal
to one divided by e raised to the power of the square root
of 25 and square root 25 simplifies to five.
And I also determine the limit as b approaches infinity,
one divided by e raise to the power of the square root of b
approaches zero because the numerator stays at one
and the denominator increases without bound.
As b approaches infinity,
the exponent on e approaches infinity,
and therefore e to the square to b also approaches infinity.
And notice one divided by e raised to the power
of the square root of 25 or one divided by e to the 1/5
is not affected by b and therefore we're left
with negative two times
negative one divided by e to the 1/5
which is equal to positive two divided by e to the 1/5.
So because the limit exists and we have a real value
for the improper integral,
we say the integral is convergent or converges.
If the limit didn't exist
because it approached positive or negative infinity
we would say the integral is divergent or diverges.
I hope you found this helpful.