Step 1: Rewrite the integrand using exponent rules. The integral is given as $\int_{x=1}^{x=2} \left(\frac{1}{x} - x^2\right) dx$. We can rewrite $\frac{1}{x}$ as $x^{-1}$. $$ \int_{1}^{2} \left(x^{-1} - x^2\right) dx $$ Step 2: Integrate each term. The integral of $x^{-1}$ is $\ln|x|$. The integral of $x^n$ is $\frac{x^{n+1}}{n+1}$. So, the integral of $-x^2$ is $-\frac{x^{2+1}}{2+1} = -\frac{x^3}{3}$. $$ \left[\ln|x| - \frac{x^3}{3}\right]_{1}^{2} $$ Step 3: Evaluate the definite integral using the limits of integration. Substitute the upper limit ($x=2$) and subtract the result of substituting the lower limit ($x=1$). $$ \left(\ln|2| - \frac{2^3}{3}\right) - \left(\ln|1| - \frac{1^3}{3}\right) $$ $$ \left(\ln(2) - \frac{8}{3}\right) - \left(\ln(1) - \frac{1}{3}\right) $$ Since $\ln(1) = 0$: $$ \left(\ln(2) - \frac{8}{3}\right) - \left(0 - \frac{1}{3}\right) $$ $$ \ln(2) - \frac{8}{3} + \frac{1}{3} $$ $$ \ln(2) - \frac{7}{3} $$ The final answer is $\boxed{\ln(2) - \frac{7}{3}}$. Send me the next one 📸