the integral in this exercise converges. evaluate the integral without using a table.\n$$\\int_{0}^{1 / 4} x…

Explanation:

Step1: Use integration by parts

The formula for integration by parts is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = \ln(4x)$ 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, $\int x\ln(4x)\mathrm{d}x=\frac{x^{2}}{2}\ln(4x)-\int\frac{x^{2}}{2}\cdot\frac{1}{x}\mathrm{d}x$. Simplify the second - integral: $\int\frac{x^{2}}{2}\cdot\frac{1}{x}\mathrm{d}x=\frac{1}{2}\int x\mathrm{d}x$.

Step2: Evaluate the remaining integral

We know that $\int x\mathrm{d}x=\frac{x^{2}}{2}+C$. So, $\frac{1}{2}\int x\mathrm{d}x=\frac{x^{2}}{4}+C$. Then $\int x\ln(4x)\mathrm{d}x=\frac{x^{2}}{2}\ln(4x)-\frac{x^{2}}{4}+C$.

Step3: Evaluate the definite integral

$\int_{0}^{\frac{1}{4}}x\ln(4x)\mathrm{d}x=\left[\frac{x^{2}}{2}\ln(4x)-\frac{x^{2}}{4}\right]{0}^{\frac{1}{4}}$. First, find the limit as $x\rightarrow0^{+}$ of $\frac{x^{2}}{2}\ln(4x)$. Use L'Hopital's rule (rewrite as $\lim{x\rightarrow0^{+}}\frac{\ln(4x)}{2x^{- 2}}$). Differentiate numerator and denominator: $\lim_{x\rightarrow0^{+}}\frac{\frac{1}{x}}{-4x^{-3}}=\lim_{x\rightarrow0^{+}}\frac{-x^{2}}{4}=0$. Now, substitute $x = \frac{1}{4}$: $\frac{(\frac{1}{4})^{2}}{2}\ln(4\times\frac{1}{4})-\frac{(\frac{1}{4})^{2}}{4}-0$. Since $\ln(1) = 0$, we have $0-\frac{1}{64}$.

Answer:

$-\frac{1}{64}$