How do you evaluate the definite integral #int1/(x^2sqrtx)# from #[1,4]#?

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Answer 1 · 2016-08-21T11:47:31

#7/12#

#int_a^b# #1/(x^2.sqrt2)# must be written as an integral of a power of x

#sqrt2#=#x^(1/2)#
So the denominator is #x ^(5/2)#

So the function can be written as #x^(-5/2)#

#int_1^4# #x^(-5/2)#.dx = [#-2/3x^(-3/2)#] from 4 to 1

When #x#=4 #x^(-3/2)#=#1/8#

#-2/3#x#1/8#=#-1/12#

When #x#=1 #x^(-3/2)# =1

#-2/3# x 1=#-2/3#

#-1/12#-(#-2/3#)=#7/12#