求 $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(3x^{2}\ln x\right)$。 由此求出 $\int x\ln x\,\mathrm{d}x$。 图中显示了直线 $x = 2$ 和曲线 $y = x\ln x$ 的一部分。该曲线与 x 轴交于点 A，与直线 $x = \tfrac{1}{2}$ 交于点 B。 求 A 点的 x 坐标。 求阴影区域的总面积。

1. Apply the product rule to differentiate the expression

Let $y = 3x^2\ln x$.
Using the product rule $\frac{\mathrm{d}}{\mathrm{d}x}(uv) = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}$ with $u = 3x^2$ and $v = \ln x$:
$\frac{\mathrm{d}}{\mathrm{d}x}\left(3x^2\ln x\right) = 3x^2 \cdot \frac{\mathrm{d}}{\mathrm{d}x}(\ln x) + \ln x \cdot \frac{\mathrm{d}}{\mathrm{d}x}(3x^2)$

2. Differentiate the standard functions and simplify

$= 3x^2 \left(\frac{1}{x}\right) + (\ln x)(6x)$
$= 3x + 6x\ln x$

1. Find the derivative from part (a)

$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}\ln x) = (6x)(\ln x) + (3x^{2})\left(\frac{1}{x}\right) = 6x\ln x + 3x$

2. Apply integration as the reverse of differentiation

$\int (6x\ln x + 3x)\,\mathrm{d}x = 3x^{2}\ln x + C$
$6\int x\ln x\,\mathrm{d}x + \int 3x\,\mathrm{d}x = 3x^{2}\ln x + C$

3. Rearrange to find the required integral

$6\int x\ln x\,\mathrm{d}x + \frac{3x^{2}}{2} = 3x^{2}\ln x + C$
$6\int x\ln x\,\mathrm{d}x = 3x^{2}\ln x - \frac{3x^{2}}{2} + C$
$\int x\ln x\,\mathrm{d}x = \frac{1}{2}x^{2}\ln x - \frac{1}{4}x^{2} + C'$

1. Set $y = 0$ to find the x-intercept

At point A, the curve meets the x-axis, so $y = 0$.
$x\ln x = 0$

2. Solve for $x$

Since $\ln x$ is only defined for $x > 0$, $x \neq 0$.
Therefore, $\ln x = 0$.
$x = e^0 = 1$

1. State the integral of $x \ln x$ from part (b)

From part (b), $\int x \ln x \, \mathrm{d}x = \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 + c$.

2. Set up the expression for the total area, splitting at the x-intercept $x=1$

Total Area = $\left| \int_{\frac{1}{2}}^{1} x \ln x \, \mathrm{d}x \right| + \int_{1}^{2} x \ln x \, \mathrm{d}x$
Total Area = $-\int_{\frac{1}{2}}^{1} x \ln x \, \mathrm{d}x + \int_{1}^{2} x \ln x \, \mathrm{d}x$

3. Evaluate the area of the region below the x-axis

Area 1 = $-\left[ \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 \right]_{\frac{1}{2}}^{1}$
$= -\left( \left( \frac{1}{2}(1)^2 \ln 1 - \frac{1}{4}(1)^2 \right) - \left( \frac{1}{2}\left(\frac{1}{2}\right)^2 \ln \frac{1}{2} - \frac{1}{4}\left(\frac{1}{2}\right)^2 \right) \right)$
$= -\left( -\frac{1}{4} - \left( \frac{1}{8}(-\ln 2) - \frac{1}{16} \right) \right)$
$= -\left( -\frac{3}{16} + \frac{1}{8}\ln 2 \right) = \frac{3}{16} - \frac{1}{8}\ln 2$

4. Evaluate the area of the region above the x-axis

Area 2 = $\left[ \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 \right]_{1}^{2}$
$= \left( \frac{1}{2}(2)^2 \ln 2 - \frac{1}{4}(2)^2 \right) - \left( -\frac{1}{4} \right)$
$= (2\ln 2 - 1) + \frac{1}{4} = 2\ln 2 - \frac{3}{4}$

5. Calculate the total area

Total Area = $\left( \frac{3}{16} - \frac{1}{8}\ln 2 \right) + \left( 2\ln 2 - \frac{3}{4} \right)$
$= \frac{15}{8}\ln 2 - \frac{9}{16}$