28. $int_{2}^{-1}3^{x}dx$

Explanation:

Step1: Recall integral formula

The integral of $a^x$ is $\frac{a^x}{\ln a}+C$. So, $\int 3^x dx=\frac{3^x}{\ln 3}+C$.

Step2: Apply fundamental theorem of calculus

$\int_{2}^{-1}3^x dx=\left[\frac{3^x}{\ln 3}\right]_{2}^{-1}=\frac{3^{-1}}{\ln 3}-\frac{3^{2}}{\ln 3}$.

Step3: Simplify the expression

$\frac{3^{-1}}{\ln 3}-\frac{3^{2}}{\ln 3}=\frac{\frac{1}{3}-9}{\ln 3}=\frac{\frac{1 - 27}{3}}{\ln 3}=-\frac{26}{3\ln 3}$.

Answer:

$-\frac{26}{3\ln 3}$