the integral in this exercise converges. evaluate the integral without using a table.\n int_{0}^{1 / 17} x…

Explanation:

Step1: Use integration by parts

Let (u = \ln(17x)), (dv=x dx). Then (du=\frac{1}{x}dx), (v=\frac{x^{2}}{2}). By integration - by - parts formula (\int_{a}^{b}u;dv=uv|{a}^{b}-\int{a}^{b}v;du), we have (\int x\ln(17x)dx=\frac{x^{2}}{2}\ln(17x)-\int\frac{x^{2}}{2}\cdot\frac{1}{x}dx). Simplify the second integral: (\int\frac{x^{2}}{2}\cdot\frac{1}{x}dx=\frac{1}{2}\int xdx=\frac{1}{4}x^{2}+C). So (\int x\ln(17x)dx=\frac{x^{2}}{2}\ln(17x)-\frac{1}{4}x^{2}+C).

Step2: Evaluate the definite integral

(\int_{0}^{\frac{1}{17}}x\ln(17x)dx=\left[\frac{x^{2}}{2}\ln(17x)-\frac{1}{4}x^{2}\right]{0}^{\frac{1}{17}}). First, find the limit as (x\rightarrow0^{+}) of (\frac{x^{2}}{2}\ln(17x)). Let (t = 17x), then (x=\frac{t}{17}) and (\lim{x\rightarrow0^{+}}\frac{x^{2}}{2}\ln(17x)=\frac{1}{2\times17^{2}}\lim_{t\rightarrow0^{+}}t^{2}\ln t). Using L'Hopital's rule (since (\lim_{t\rightarrow0^{+}}t^{2}\ln t=\lim_{t\rightarrow0^{+}}\frac{\ln t}{t^{- 2}}), and (\lim_{t\rightarrow0^{+}}\frac{\ln t}{t^{-2}}=\lim_{t\rightarrow0^{+}}\frac{\frac{1}{t}}{-2t^{-3}}=\lim_{t\rightarrow0^{+}}\frac{-t^{2}}{2}=0)). When (x = \frac{1}{17}), (\frac{(\frac{1}{17})^{2}}{2}\ln(17\times\frac{1}{17})-\frac{1}{4}(\frac{1}{17})^{2}). Since (\ln(1) = 0), the first term is (0). So (\int_{0}^{\frac{1}{17}}x\ln(17x)dx=-\frac{1}{4\times17^{2}}=-\frac{1}{1156}).

Answer:

(-\frac{1}{1156})