Evaluate the indefinite integral.
${\int }_{}^{}2{\left(\frac{1}{3}\right)}^{x}dx$

• A. $-\frac{2}{3^x\ln 3}+C$
• B. $-\frac{2x^{x+1}}{4}+C$
• C. $-\frac{2x}{\ln 3}+C$
• D. $-\frac{2x^{x+1}}{3^x\ln 4}+C$

Answer: (A)

Verified step by step guidance

1. Step 1: Recognize the integral to be evaluated: ∫2(1/3)^x dx. This is an exponential function with a base of 1/3. Recall that the general formula for integrating an exponential function of the form ∫a^x dx is (a^x / ln(a)) + C, where a > 0 and a ≠ 1.
2. Step 2: Rewrite the given integral to make it easier to work with. The integral can be expressed as ∫2 * (1/3)^x dx. Here, the constant 2 can be factored out of the integral, giving 2 * ∫(1/3)^x dx.
3. Step 3: Apply the formula for integrating exponential functions. For the term (1/3)^x, the base is 1/3. Using the formula, the integral of (1/3)^x is (1/3)^x / ln(1/3).
4. Step 4: Simplify the result. Note that ln(1/3) = ln(1) - ln(3) = -ln(3). Substituting this into the expression, the integral becomes -((1/3)^x / ln(3)). Don't forget to multiply by the constant 2 factored out earlier.
5. Step 5: Add the constant of integration, C, to account for the indefinite integral. The final expression is -2 * ((1/3)^x / ln(3)) + C.