Question

In Exercises \(27-40\) , evaluate each integral using Part 2 of the Fundamental Theorem. Support your answer with NINT if you are unsure. $$\int_{2}^{-1} 3^{x} d x$$

Short answer

The result of the integral is \( \frac{1}{3\ln(3)} - \frac{9}{\ln(3)} \)

Step-by-step solution

Find the antiderivative of the function

Our function to integrate is \(3^x\). The antiderivative of \(a^x\) where \(a>0\), is \(\frac{a^x}{\ln(a)}\). So the antiderivative of \(3^x\) is \(\frac{3^x}{\ln(3)}\).

Apply the Fundamental Theorem of Calculus

Substitute the limits of integration into the antiderivative. This means replacing every \(x\) in \(\frac{3^x}{\ln(3)}\) with the upper limit \(-1\) first and then the lower limit \(2\), and subtract the latter from the former, i.e., \[ \frac{3^{-1}}{\ln(3)} - \frac{3^2}{\ln(3)} \]

Compute the result

Simplify the elements in the subtraction, to get the final result, \[ \frac{1}{3\ln(3)} - \frac{9}{\ln(3)} \]