Find area of region bounded by y = (25(3-x)(x-6))/(x^2) and

[val1]

I need to find the area of the region which is bounded by the x-axis and by the graph of \(\displaystyle \L y = \frac{{25(3 - x)(x - 6)}}{{x^2 }}\). Please give the answer to two decimal places.

I'm a bit stuck on this one. Thank you

 

[galactus]

Can you see what the zeros are by looking at the function?. That will determine your limits of integration. It crosses the x-axis at 3 and 6. See?. Expanding out may make it easier to integrate.

\(\displaystyle \L\\\frac{25(3-x)(x-6)}{x^{2}}=\frac{-450}{x^{2}}+\frac{225}{x}-25\)

Can you set it up now?.

 

[val1]

Thanks for your reply, Galactus

Here's where I have got to:

\(\displaystyle \L \int\limits_3^6 { - 450x^{ - 2} + } 225\int\limits_3^6 {\frac{1}{x} - \int\limits_3^6 {25} } = 450[x^{ - 1} ]_3^6 + 225[\ln (x)]_3^6 - 25[x]_3^6 ]\)

\(\displaystyle \L = - 75 + 155.9581 - 75 = 5.96\)

Am I on the right track?

 

[galactus]

That's correct. To be anal, perhaps leave your answer as

\(\displaystyle \L\\225ln(2)-150 \;\ or \;\ 75(3ln(2)-2)\). Looks better. Good work.