What is the net area between #f(x)=x/ln(x^2)# in #x in[3,10] # and the x-axis?

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Answer 1 · 2018-05-06T15:50:33

The area equal
#A=int_3^10x/ln(x^2)*dx=[(Gamma(0,-2*ln(x)))/2]_3^10=[(Gamma(0,-2*ln(3))-Gamma(0,-2*ln(10)))/2]=12.20245(unite)^2#

The sketch of our function #f(x)=x/ln(x^2)#

graph{x/ln(x^2) [-10, 10, -5, 5]}

the interval #x in[3,10]#

enter image source here

this is the area wanted

#A=int_a^by*dx#

#A=int_3^10x/ln(x^2)*dx=[(Gamma(0,-2*ln(x)))/2]_3^10=[(Gamma(0,-2*ln(3))-Gamma(0,-2*ln(10)))/2]=12.20245(unite)^2#