The problem I must solve for a high-school math paper is the value of the definite integral:
$$\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8x^3)}{kx}dx, k\in \mathbb{R}^+$$
Not looking too hard at first glance, I solved it as such:
$$\begin{align}
\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8x^3)}{kx}dx
&=\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8)+3\ln(x)}{kx}dx\\
&=\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8)}{kx}dx+\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{3\ln(x)}{kx}dx\\ &=(\frac{\ln(8)\ln(x)}{k} +\frac{3\ln^2(x))}{2k} ) \Big|_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\\
&=(\frac{\ln(8)}{2k} +\frac{3((\frac{3}{2}-\ln(2))^2-(\frac{1}{2}-\ln(2))^2)}{2k} ) \\
&=\frac{6-3\ln(2)}{2k}
\end{align}$$
However, the "technically correct" way of solving the integral is by direct u-substitution.
This is done by setting $u=\ln(8x^3)$ and then $du=\frac{3}{x}dx$. Then $\frac{1}{kx}dx = \frac{1}{3k}du$ and hence:
$$\int_{\frac{e}{2}}^{\frac{e^{3/2}}{2}}\frac{\ln(8x^3)}{kx}dx=\frac{1}{3k}\int_{3}^{\frac{9}{2}}udu=\frac{15}{8k}$$
These results are obviously distinct from each other, as $4\times(6-3\ln(2)) \neq 15$. It confuses me why my method, although different from the expected method, obtains the incorrect answer. I also can't seem to find any mistakes in my own working out.
Any help will be greatly appreciated. Thanks!
Best Answer
It turns out I have made a mistake.
$$(\frac{\ln(8)}{2k} +\frac{3((\frac{3}{2}-\ln(2))^2-(\frac{1}{2}-\ln(2))^2)}{2k} ) $$
This step is incorrect as $\ln(\frac{e}{2})=1-\ln(2)$, rather than $\frac{1}{2}-\ln(2)$. As a result, the definite integral is not evaluated correctly. Using the correct expression:
$$(\frac{\ln(8)}{2k} +\frac{3((\frac{3}{2}-\ln(2))^2-(1-\ln(2))^2)}{2k} ) $$
The terms for $\ln(2)$ cancel out and, indeed, we are left with only $\frac{15}{8k}$