How do you use the integral test to determine whether #int lnx/(xsqrtx)# converges or diverges from #[3,oo)#?
1 Answer
The integral converges to
Explanation:
We won't use the "integral test," since that test uses integrals to determine whether or not certain series converge or diverge. We will, however, integrate the given function and evaluate it at its infinite limit and
#intlnx/(xsqrtx)dx=intx^(-3/2)lnxdx#
Let's use integration by parts. Let:
#{(u=lnx" "=>" "du=1/xdx),(dv=x^(-3/2)dx" "=>" "v=-2x^(-1/2)):}#
So the (presently unbounded) integral becomes:
#=-2x^(-1/2)lnx-int-2x^(-1/2)(1/xdx)#
#=-(2lnx)/sqrtx+2intx^(-3/2)dx#
As we've already determined:
#=-(2lnx)/sqrtx-4/sqrtx#
#=-(2(lnx+2))/sqrtx#
Thus, the improper integral is:
#int_3^oolnx/(xsqrtx)dx=-2[(lnx+2)/sqrtx]_3^oo#
#=-2lim_(xrarroo)(lnx+2)/sqrtx-(-2)(ln3+2)/sqrt3#
The limit is
#=2/sqrt3(ln3+2)#
The integral converges.
