Step 1: Write the given definite integral. $$ \int_{x=1}^{x=2} \left(x^2 - \frac{1}{x}\right) dx $$ Step 2: Find the antiderivative of each term in the integrand. The antiderivative of $x^2$ is $\frac{x^{2+1}}{2+1} = \frac{x^3}{3}$. The antiderivative of $\frac{1}{x}$ is $\ln|x|$. So, the antiderivative of $\left(x^2 - \frac{1}{x}\right)$ is $\frac{x^3}{3} - \ln|x|$. Step 3: Evaluate the definite integral using the Fundamental Theorem of Calculus. $$ \left[\frac{x^3}{3} - \ln|x|\right]_{1}^{2} $$ Substitute the upper limit ($x=2$) and subtract the result of substituting the lower limit ($x=1$). $$ \left(\frac{2^3}{3} - \ln|2|\right) - \left(\frac{1^3}{3} - \ln|1|\right) $$ Step 4: Simplify the expression. $$ \left(\frac{8}{3} - \ln 2\right) - \left(\frac{1}{3} - 0\right) $$ Since $\ln 1 = 0$. $$ \frac{8}{3} - \ln 2 - \frac{1}{3} $$ Combine the fractional terms. $$ \left(\frac{8}{3} - \frac{1}{3}\right) - \ln 2 $$ $$ \frac{7}{3} - \ln 2 $$ The final answer is: $$\boxed{\frac{7}{3} - \ln 2}$$ That's 2 down. 3 left today — send the next one.