Evaluate the indefinite integral.∫2(13)xdx\int_{}^{}2\left(\frac13\right)^{x}dx∫2(31)xdx

− 2 3 x ln ⁡ 3 + C -\frac{2}{3^{x}\ln3}+C − 3 x l n 3 2 + C

Evaluate the indefinite integral. ∫ 2 ( 1 3 ) x d x \int_{}^{}2\left(\frac13\right)^{x}dx ∫ 2 ( 3 1 ) x d x

− 2 3 x ln ⁡ 3 + C -\frac{2}{3^{x}\ln3}+C − 3 x l n 3 2 + C

− 2 x x + 1 4 + C -\frac{2x^{x+1}}{4}+C − 4 2 x x + 1 + C

− 2 x ln ⁡ 3 + C -\frac{2x}{\ln3}+C − l n 3 2 x + C

− 2 x x + 1 3 x ln ⁡ 4 + C -\frac{2x^{x+1}}{3^{x}\ln4}+C − 3 x l n 4 2 x x + 1 + C

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

Business Calculus

10. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals of Exponential Functions

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{}2\left(\frac13\right)^{x}dx$

$-\frac{2}{3^{x}\ln3}+C$

$-\frac{2x^{x+1}}{4}+C$

$-\frac{2x}{\ln3}+C$

$-\frac{2x^{x+1}}{3^{x}\ln4}+C$

Verified step by step guidance

Step 1: Recognize that the integral involves an exponential function of the form $ \left( \frac{1}{3} \right)^x $. Rewrite this as $ 2 \cdot \left( \frac{1}{3} \right)^x $, which can also be expressed as $ 2 \cdot 3^{-x} $.

Step 2: Recall the general formula for integrating exponential functions: $ \int a^{u} \cdot u' \, dx = \frac{a^{u}}{\ln(a)} + C $, where $ a > 0 $ and $ a \neq 1 $. In this case, $ a = 3 $ and $ u = -x $.

Step 3: Differentiate $ u = -x $ to find $ u' = -1 $. This means we need to account for the negative sign when integrating.

Step 4: Apply the formula for the integral. The integral becomes $ \int 2 \cdot 3^{-x} \, dx = 2 \cdot \frac{3^{-x}}{-\ln(3)} + C $. Simplify the expression to $ -\frac{2}{3^x \ln(3)} + C $.

Step 5: Combine the constant of integration $ C $ to complete the solution. The final expression is $ -\frac{2}{3^x \ln(3)} + C $.

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