Not really sure what you want to do. What you have done is find where y is given by both
\(\displaystyle y = 5x^{-1}\)
and
\(\displaystyle y = 10x^{-2}\)
which, as you have shown with another minor step, would give
x = 2
But that leaves the x = 7 hanging out there to dry.
7 isn't really out there to dry. It's a given, and it's not an intersection of the two lines.
\(\displaystyle y = 5x^{-1}\)
\(\displaystyle y = 10x^{-2}\)
\(\displaystyle x = 7\)
Finding the other limit of integration:
\(\displaystyle 5x^{-1} = 10x^{-2}\)
Next Step?
\(\displaystyle (5x^{-1})^{-1} = (10x^{-2})^{-1}\)
\(\displaystyle \dfrac{1}{5}x = \dfrac{1}{10}x^{2}\)
\(\displaystyle x = \dfrac{5}{10}x^{2}\)
\(\displaystyle x = \dfrac{1}{2}x^{2}\)
\(\displaystyle 1= \dfrac{1}{2}x\)
\(\displaystyle 2 = x\)
(less steps - better way) Another strategy would be to get one single x value on one side, and then (using power manipulation) make that into \(\displaystyle x^{1}\)
\(\displaystyle y = 5x^{-1}\)
\(\displaystyle y = 10x^{-2}\)
\(\displaystyle x = 7\)
Finding the other limit of integration:
\(\displaystyle 5x^{-1} = 10x^{-2}\)
\(\displaystyle 5 = 10x^{-1}\)
\(\displaystyle (5)^{-1} = (10x^{-1})^{-1}\)
\(\displaystyle \dfrac{1}{5} = \dfrac{1}{10}x\)
\(\displaystyle \dfrac{10}{5} = x\)
\(\displaystyle 2 = x\)
Here is another way:
Get the variable on one side, without a constant attached, then make it into \(\displaystyle x^{-1}\)
\(\displaystyle y = 5x^{-1}\)
\(\displaystyle y = 10x^{-2}\)
\(\displaystyle x = 7\)
Finding the other limit of integration:
\(\displaystyle 5x^{-1} = 10x^{-2}\)
\(\displaystyle 5 = 10x^{-1}\)
\(\displaystyle \dfrac{5}{10} = x^{-1}\)
\(\displaystyle (\dfrac{5}{10})^{-1} = (x^{-1})^{-1}\)
\(\displaystyle (\dfrac{1/5}{1/10})= x\)
\(\displaystyle 2 = x\)
Now integrating:
\(\displaystyle \int_{2}^{7} 5x^{-1} - 10x^{-2} dx\)
\(\displaystyle \dfrac{1}{5}\ln|5x| + 10x^{-1}\) evaluated at 7 and 2
\(\displaystyle [\dfrac{1}{5}\ln|5(7)| + 10(7)^{-1}] - [\dfrac{1}{5}\ln|5(2)| + 10(2)^{-1}] = -3.320875979\) ??
