How do you evaluate the integral #int x dx# from #-oo# to #oo# if it converges?

1 Answer
Aug 21, 2016

#0#

Explanation:

We have:

#int_(-oo)^ooxdx#

Integrating:

#=[x^2/2]_(-oo)^oo#

We can't technically "plug in" infinity and negative infinity, so take their limits:

#=lim_(xrarroo)(x^2/2)-lim_(xrarr-oo)(x^2/2)#

Both approach positive infinity:

#=oo-oo=0#

This should make sense, if we think about #int_(-oo)^ooxdx# as describing the area between the line #y=x# and the #x#-axis. The negative portion where #x<0# is completely balanced out by the positive area where #x>0#:

graph{x [-304.4, 304.4, -152.2, 152.2]}