The question states integrate
x log(yz) + log(xy)/z
with bounds 1 < x < 2, 1 < y < 2, 1 < z < 2.
Really do not know how to go about solving this problem
Not so sure which order of integration is best to choose either.
I'm thinking it's better if I separate the integral out into two separate ones then integrate them separately. If I do that could I use different orders of integration for the two seperate integrals. Solve the integral and then just add it???
so I would have
integral(xlog(yz)) + integral(log(xy)/z)
then for each separate integral integrate them separately by parts??
If someone could help me solve this problem I would really appreciate it.
x log(yz) + log(xy)/z
with bounds 1 < x < 2, 1 < y < 2, 1 < z < 2.
Really do not know how to go about solving this problem
Not so sure which order of integration is best to choose either.
I'm thinking it's better if I separate the integral out into two separate ones then integrate them separately. If I do that could I use different orders of integration for the two seperate integrals. Solve the integral and then just add it???
so I would have
integral(xlog(yz)) + integral(log(xy)/z)
then for each separate integral integrate them separately by parts??
If someone could help me solve this problem I would really appreciate it.
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∫ ln(kx) dx = x (ln(kx) - 1) + c ..... needed several times for this problem
∫ ∫ ∫ x ln(yz) + ln(xy)/z dx dy dz
{(x,y,z) | 1 ≤ x ≤ 2 , 1 ≤ y ≤ 2 , 1 ≤ z ≤ 2}
= ∫ ∫ x(2 ln(2y) - ln(y) - 1) + ln(2) ln(xy) dx dy
{(x,y) | 1 ≤ x ≤ 2 , 1 ≤ y ≤ 2}
= ∫ ln(2) (2 ln(2x) - ln(x) - 1) + x (ln(16) - 2) dx
{(x) | 1 ≤ x ≤ 2}
= ln(16) (ln(2) + 1) - 3 ≈ 1.6944
Answer: ln(16) (ln(2) + 1) - 3 ≈ 1.6944
∫ ∫ ∫ x ln(yz) + ln(xy)/z dx dy dz
{(x,y,z) | 1 ≤ x ≤ 2 , 1 ≤ y ≤ 2 , 1 ≤ z ≤ 2}
= ∫ ∫ x(2 ln(2y) - ln(y) - 1) + ln(2) ln(xy) dx dy
{(x,y) | 1 ≤ x ≤ 2 , 1 ≤ y ≤ 2}
= ∫ ln(2) (2 ln(2x) - ln(x) - 1) + x (ln(16) - 2) dx
{(x) | 1 ≤ x ≤ 2}
= ln(16) (ln(2) + 1) - 3 ≈ 1.6944
Answer: ln(16) (ln(2) + 1) - 3 ≈ 1.6944