Welcome to an example of an improper integral, where
we have a discontinuous integrand
on the interval of integration.
We want to determine if the integral converges or diverges.
If it converges, we want to evaluate the integral.
And if it diverges, we want to indicate if it approaches
positive or negative infinity.
So looking at the integral, notice
how it might not seem like an improper integral
because the lower limit of integration
is not negative infinity or the upper limit of integration
is not positive infinity.
But if we take a look at the integrand here,
notice how when x equals 5, we'd have a denominator of 0,
meaning we'd have division by 0.
And, therefore, the function value is undefined.
And, therefore, we do have an improper integral.
We could also verify this graphically.
Notice how the graph of the integrand
would be this rational function where
we have a vertical asymptote at x equals positive 5.
So if we want to integrate from 5 to 8--
well, if they write it as a limit approaching 5
from the positive side or right side--
so we'll write this as the limit of,
let's say, c approaches 5 from the positive side.
And then we'll substitute c for 5,
giving us the integral from c to 8.
Let's rewrite the integrand as 2 times the quantity x minus 5
to the negative 1/3.
Remember, if we have a cube root,
it's equivalent to a to the 1/3 power.
And if we have 1 divided by a to the positive 1/3 power,
that's equivalent to a to the negative 1/3 power.
So we applied both of these properties
here to rewrite the integral.
And now we'll integrate.
We might be thinking u-substitution here,
where we let u equal the quantity x minus 5.
If we do this, notice how du is just equal to dx.
So we can go ahead and just add 1 to the exponent
and then divide by the new exponent.
So we'd have the limit as c approaches 5
from the right of 2 times the quantity x minus 5
to the positive 2/3 power divided by 2/3.
Our limits of integration are from c to 8.
But instead of dividing by 2/3, we
can multiply by the reciprocal of 3/2.
2 times 3/2 equals 3.
So we can write this as limit as c approaches 5
from the right of 3 times the quantity x minus 5 to the 2/3.
And I will first substitute 8 for x, and then c for x.
So when x is 8, we would have 3 times--
8 minus 5 is 3-- so 3 to the 2/3 minus--
substituting c for x, we would have 3 times the quantity
c minus 5 raised to the 2/3 power.
And now we'll determine this one-sided limit.
Notice how this first term is not affected by c at all.
So we would just have 3 times 3 to the 2/3 minus--
as c approaches 5 from the right, c minus 5 approaches 0.
So this term approaches 0.
And, therefore, the limit is just 3 times 3
to the 2/3 power.
Now, if we wanted to, in radical form, this would be 3 times--
the denominator would be the index, the cube
root of 3 squared, or 9.
So we can see the limit does exist.
And, therefore, the integral does converge.
And its value is 3 times 3 to the 2/3 power,
or 3 times the cube root of 9.
And just in case you're wondering,
this would be approximately 6.2403
if we wanted a decimal approximation.
Now, if we go back and take a look
at the graph of this function again on this interval, notice
how it is non-negative, which means on the interval
from 5 to 8, the area below the function and above the x-axis
would be 3 times the cube root of 9 square units.
So the area of this shaded region--
it's exactly 3 times the cube root of 9 square units,
or we can say it's approximately 6.2403 square units.
I hope you found this helpful.