We want to evaluate the given definite integral
using integration by parts.
The integration by parts formula is here.
It states the integral of u dv equals u times v
minus the integral of v du.
So the integral of u dv is difficult to integrate,
and hopefully after applying this formula,
the integral of v du will be easier to integrate.
The first thing we should do is move x to the negative two
up to the numerator which changes the sign of the exponent
giving us the integral of x to the power of negative two
times natural log x dx.
We want to select u so that differential u is simpler.
We want to select dv so that we can easily integrate
to find v.
So in this case, let's try that in u equal natural log x.
Since natural log x is not easy to integrate,
it wouldn't be a good choice for dv.
So if we let u equal natural log x,
then that means differential v or dv
would be equal to x to the negative two dx.
So if we let u equal natural log x,
then differential u or du is equal to u prime
times dx, which would be one divided by x dx.
And if we let dv equal x to the power of negative two dx,
we would integrate both sides to find v.
So if we integrate x to the power of negative two,
we would add one to the exponent
and then divide by the new exponent.
That would give us x to the power of negative one
divided by negative one
or just negative x to the negative one.
Now we can apply the integration by parts formula.
So this given integral is equal to u times v.
Well here's u and here's v.
So natural log x times negative x to the negative one,
would be equivalent to negative one divided by
x natural log x.
And we do have a definite integral.
So we'll have to evaluate this at four and one
and then find the difference.
And then we still have minus the integral of v du.
So here's v and here's du.
So we'd have negative x to the negative one
times one over x dx.
And let's write this as x to the negative one dx.
Of course, the limits of integration
are still from one to four here.
And then we have minus and negative that'll be plus
the integral of, and here we're multiplying,
so we add the exponent,
so we'll have x to the power of negative two dx.
So we have negative one divided by x natural log x
plus the entire derivative of x to the negative two.
We add one to the exponent
that'll give us x to the negative one,
and then we divide by negative one,
so we'll have negative x to the negative one.
And we'll evaluate this at four then one
then find the difference.
Let's finish this on the next slide.
Let's go ahead and write this as
negative one divide by x, natural log x.
And then let's write this as minus one over x.
So when x is four we'd have negative 1/4 natural log four
minus 1/4,
and then when x is one we have negative one over one,
which is negative one times natural log one minus
one over one which is one,
which give us negative 1/4 natural log four minus 1/4.
Natural log one is equal to zero.
So this simplifies out.
So we just have minus negative one or plus one.
So the exact value would be negative 1/4 natural log four,
and then negative 1/4 plus one would be plus 3/4.
Let's also get our decimal approximation for this.
We'd have negative 1/4,
natural log four,
and then plus 3/4.
So we have approximately 0.4034.
And because we had a definite integral,
we can actually check this on the graphing calculator.
To do this, we'll press MATH, option nine
for function integration,
and this is the newer version of the TI-84,
so if you have an older version,
it will look a little bit different.
If I press Enter, notice how we see an actual integral here.
So we first enter the lower limit of integration
which is one, right arrow,
upper limit of integration is four, right arrow.
Let's enter the integrand in this form here.
So we have x raised to the power of negative two,
right arrow to exit the exponent,
and then we have natural log x, closed parenthesis,
right arrow, the variable of integration which is x,
and Enter.
And notice how this does verify our result,
as we see here as our decimal approximation.
I hope you found this helpful.