Can someone please check my solution and see where my mistake is (1 Viewer)
- Thread starter kpad5991
- Start date
When x = 2 y= 0 when y = 1-2/x.. Based on that graph, the desired equation may have been $y = 1 -\frac{2}{x}$ instead.$)
Is the solution 2+ log2-2log 3?
How is the solution 2-1/2log 3???
When I changed the equation I still didn't get the textbook solutions
dan964
what
Area is a positive value, so split the integral. Using the graph rather than the equation (assuming typo):
\right| + \left|\displaystyle\int_{2}^{3}(1-\frac{2}{x}\; dx)\right| \\ &= |[x-2 \ln x]^{2}_{1}| + |[x-2 \ln x]^{3}_{2}|\\ &= |1- 2\ln 2| + |1 - 2\ln 3 + 2 \ln 2| \\ &= 4 \ln 2 - 2 \ln 3 \end{align*} )
Note: the coefficients would be changed to
if using the equation given rather than the graph.
If not finding area, removing the absolute value signs gives:
(using graph) or
(using equation)
Note: the coefficients would be changed to
if using the equation given rather than the graph.
If not finding area, removing the absolute value signs gives:
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dan964
what
so it seems it is a poorly worded question and the diagram is just wrong/typo in question.
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I get all the steps except how do you go fromArea is a positive value, so split the integral. Using the graph rather than the equation (assuming typo):
Note: the coefficients would be changed to
if using the equation given rather than the graph.
If not finding area, removing the absolute value signs gives:
4ln 2 -2ln 3 to 2-1/2ln3???!
dan964
what
ah ok. YOU DON'T, THEY ARE ANSWERS TO DIFFERNT PROBLEMS.
The first 2 answers is when you find area, considering it is as a postive quantity. Split the integral and use | . | signs.
The second 2 answers is when you evaluate the integral (sometimes called 'area' as well erroneously I suppose). The textbook answer does this. But I don't think you can say that is right, but the question isn't clear.
The equation of the graph is
Here is the working, note no need for 2 integrals:
The 1/2 comes if the original graph is y=1/2 x.
The answer in the book is wrong; or the question is wrong. Either way to get the exact answer
What is more important is can you explain why the textbook answer is wrong. Whole point of MX2 requires higher level thinking.
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