How to solve this?#f:RR#\{2}#->RR;f(x)=x^2/(x-2)#.Demonstrate that #8<=int_3^4f(x)dx<=9#

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To solve this problem, we'll first find the integral of ( f(x) ) over the interval ([3, 4]), and then demonstrate that the result falls within the given bounds of ( 8 \leq \int_{3}^{4} f(x) , dx \leq 9 ).

Let's start by finding the integral:

[ \int_{3}^{4} \frac{x^2}{x - 2} , dx ]

We can perform a polynomial division to simplify the integrand:

[ \frac{x^2}{x - 2} = x + 2 + \frac{4}{x - 2} ]

Now, integrate term by term:

[ \int_{3}^{4} x , dx + \int_{3}^{4} 2 , dx + \int_{3}^{4} \frac{4}{x - 2} , dx ]

[ = \frac{x^2}{2} + 2x + 4 \ln|x - 2| \bigg|_{3}^{4} ]

[ = \frac{16}{2} + 2(4) + 4 \ln|4 - 2| - \left(\frac{9}{2} + 2(3) + 4 \ln|3 - 2|\right) ]

[ = 8 + 8 + 4 \ln 2 - \left(\frac{9}{2} + 6 + 4 \ln 1\right) ]

[ = 16 + 4 \ln 2 - \left(\frac{9}{2} + 6\right) ]

[ = 16 + 4 \ln 2 - \frac{21}{2} ]

Now, calculate the exact numerical value:

[ \approx 8.3863 ]

Since ( 8 \leq \int_{3}^{4} f(x) , dx \leq 9 ) is given, and ( 8.3863 ) falls within this range, we've demonstrated that ( 8 \leq \int_{3}^{4} f(x) , dx \leq 9 ).