Question

Compute the value of the following improper integral if it converges. If it diverges, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise (hint: integrate by parts). ∫∞18ln(x)x2dx Determine whether ∑n=1∞(8ln(n)n2) is a convergent series. Enter C if the series is convergent, or D if it is divergent.

Answer 1

To compute the value of the improper integral, we can integrate by parts. Using the formula for integration by parts, we find that the integral converges to a finite value of -ln(x)/x as x approaches infinity.

Step-by-step explanation:

To compute the value of the improper integral ∫∞18ln(x)/x2dx, we can integrate by parts. Let u = ln(x) and dv = 1/x2dx. Differentiating u with respect to x gives du = 1/x dx and integrating dv gives v = -1/x. Applying the formula for integration by parts, we get:

∫∞18ln(x)/x2dx = -ln(x)/x + ∫∞181/x2dx.

Simplifying the integral, we have:

∫∞181/x2dx = -1/x

As x approaches infinity, 1/x approaches 0. Therefore, the improper integral converges to a finite value of -ln(x)/x.

Answer 2

INF for first while D for second

Step-by-step explanation:

Ok I think I read that integral with lower limit 1 and upper limit infinity

where the integrand is ln(x)*x^2

integrate(ln(x)*x^2)

=x^3/3 *ln(x)- integrate(x^3/3 *1/x)

Let's simplify

=x^3/3 *ln(x)-integrate(x^2/3)

=x^3/3*ln(x)-1/3*x^3/3

=x^3/3* ln(x)-x^3/9+C

Now apply the limits of integration where z goes to infinity

[z^3/3*ln(z)-z^3/9]-[1^3/3*ln(1)-1^3/9]

[z^3/3*ln(z)-z^3/9]- (1/9)

focuse on the part involving z... for now

z^3/9[ 3ln(z)-1]

Both parts are getting positive large for positive large values of z

So the integral diverges to infinity (INF)

By the integral test... the sum also diverges (D)