the integral in this exercise converges. evaluate the integral without using a table.\n int_{0}^{1 / 20} x…

Explanation:

Step1: Use integration by parts

Integration by parts formula: $\int u\mathrm{d}v=uv - \int v\mathrm{d}u$. Let $u = \ln(20x)$ and $\mathrm{d}v=x\mathrm{d}x$. Then $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$ and $v=\frac{x^{2}}{2}$. So, $\int x\ln(20x)\mathrm{d}x=\frac{x^{2}}{2}\ln(20x)-\int\frac{x^{2}}{2}\cdot\frac{1}{x}\mathrm{d}x=\frac{x^{2}}{2}\ln(20x)-\frac{1}{2}\int x\mathrm{d}x$.

Step2: Evaluate the remaining integral

$\int x\mathrm{d}x=\frac{x^{2}}{2}+C$. Then $\int x\ln(20x)\mathrm{d}x=\frac{x^{2}}{2}\ln(20x)-\frac{x^{2}}{4}+C$.

Step3: Evaluate the definite integral

$\lim_{a\rightarrow0^{+}}\int_{a}^{\frac{1}{20}}x\ln(20x)\mathrm{d}x=\left[\frac{x^{2}}{2}\ln(20x)-\frac{x^{2}}{4}\right]{a}^{\frac{1}{20}}$. When $x = \frac{1}{20}$: $\frac{(\frac{1}{20})^{2}}{2}\ln(20\times\frac{1}{20})-\frac{(\frac{1}{20})^{2}}{4}=\frac{1}{800}\ln(1)-\frac{1}{1600}=-\frac{1}{1600}$. When $x=a$: $\lim{a\rightarrow0^{+}}\left(\frac{a^{2}}{2}\ln(20a)-\frac{a^{2}}{4}\right)$. Use L - H rule for $\lim_{a\rightarrow0^{+}}a^{2}\ln(20a)=\lim_{a\rightarrow0^{+}}\frac{\ln(20a)}{\frac{1}{a^{2}}}$. By L - H rule (since $\lim_{a\rightarrow0^{+}}\ln(20a)=-\infty$ and $\lim_{a\rightarrow0^{+}}\frac{1}{a^{2}}=\infty$), $\lim_{a\rightarrow0^{+}}\frac{\ln(20a)}{\frac{1}{a^{2}}}=\lim_{a\rightarrow0^{+}}\frac{\frac{1}{a}}{-\frac{2}{a^{3}}}=\lim_{a\rightarrow0^{+}}(-\frac{a^{2}}{2}) = 0$. And $\lim_{a\rightarrow0^{+}}\frac{a^{2}}{4}=0$.

Answer:

$-\frac{1}{1600}$