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find the area of the region under y=6 ln(3 x) and above y=8 for 4<x<7

3 answers

Integrate for x=4 to 7 the vertical slices
from y2(x) to y1(x).

y1(x)=6ln(3x)
y2(x)=8

so
I=∫[6ln(x)-8]dx from 4 to 7

I get 6.46 approx.

I got 26.23

Did you take out the 3 from ln(3x)?

Yes, your answer is correct.
I missed out the three.

Ask a New Question or answer this question.

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