How to integrate #int_0^1x^2e^(4x) dx# ?

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Answer 1 · 2018-05-11T15:09:39

#I=1/32(5e^4-1)#

Here,

#I=int_0^1 x^2e^(4x)dx#

#"Using "color(blue)"Integration by Parts:"#

#I=[x^2 ((e^(4x))/4)]_0^1 -int_0^1(2x)((e^(4x))/4)dx#

#I=[1*(e^4/4)-0]-1/2int_0^1xe^(4x)dx#

Again, #"Using "color(blue)"Integration by Parts:"# in second fraction

#I=1/4e^4-1/2[x*((e^(4x))/4)]_0^1 +1/2int_0^1(e^(4x)/4)dx#

#I=1/4e^4-1/2[1*(e^4/4)]+1/8[(e^(4x)/4)]_0^1#

#=1/4e^4-1/8e^4+1/8[e^4/4-e^0/4]#

#=1/4e^4-1/8e^4+1/32e^4-1/32#

#=1/32e^4(8-4+1)-1/32#

#I=5/32e^4-1/32#

#I=1/32(5e^4-1)#