evaluate the definite integral.\n int_{3}^{5} \frac{4}{(2 x-5)^{2}} d x \n int_{3}^{5} \frac{4}{(2 x-5)^{2}}…

Explanation:

Step1: Substitution

Let (u = 2x - 5), then (du=2dx), and (dx=\frac{1}{2}du). When (x = 3), (u=2\times3 - 5=1). When (x = 5), (u=2\times5 - 5 = 5). The integral (\int_{3}^{5}\frac{4}{(2x - 5)^{2}}dx) becomes (\int_{1}^{5}\frac{4}{u^{2}}\times\frac{1}{2}du). Simplify the integrand: (\int_{1}^{5}\frac{2}{u^{2}}du=2\int_{1}^{5}u^{- 2}du).

Step2: Integrate using the power rule

The power rule for integration is (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C) ((n\neq - 1)). For (n=-2), (\int u^{-2}du=\frac{u^{-2 + 1}}{-2+1}+C=-\frac{1}{u}+C). So (2\int_{1}^{5}u^{-2}du=2\left[-\frac{1}{u}\right]_{1}^{5}).

Step3: Evaluate the definite integral

Using the fundamental theorem of calculus (\left.F(x)\right|_{a}^{b}=F(b)-F(a)). (2\left(-\frac{1}{5}+\frac{1}{1}\right)=2\left(\frac{-1 + 5}{5}\right)=2\times\frac{4}{5}=\frac{8}{5}).

Answer:

(\frac{8}{5})