Find the area of the region enclosed between the two curves y = e0 ln and y = 1/

Question

Find the area of the region enclosed between the two curves y = e0 ln and y = 1/x on [3,4] The area of the region is square units. (Do not round until the final answer. Then round to three decimal place as needed.) x11(1 + x12)12 dx x11 (1 - x12)12 dx = (Type any exact answer.)

Explanation / Answer

(1) area enclosed by the given two curves = modulus of [area of curve 1-area of curve 2]under limits 3 to4. =integral of[e^.6x-1/x]dx under limits 3 and 4 =[e^.6x/.6x-ln(x)]under limits 3 and 4 =[e^2.4/2.4-ln(4)]-[e^1.8/1.8-ln(3)] =1.235-.28 =.955 units (2) assume a variable t=1-x^12 =>dt=(-12x^11)dx=>dx=(-dt/(12t^11)) now we transform the given integral in the form of variable t it will be transformed as integral of[-t^11dt/12] now here note that we should change the limits, we should take the limits of t at x=0 , t=1, at x=1,t=0. so our limits will be 1 to 0 integral of[-t^11dt/12] under limits 1 to 0=(-t^12/144) under limits 1 to 0 =[-(-1/144)]=1/144

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