Question 4

Let the integral be $I = \int_0^2 \frac{1}{3^x + 3} dx$.

To solve this, we can divide the numerator and the denominator by $3^x$: $I = \int_0^2 \frac{3^{-x}}{1 + 3 \cdot 3^{-x}} dx$

Now, let $u = 1 + 3 \cdot 3^{-x}$. Differentiating both sides with respect to $x$: $\frac{du}{dx} = 3 \cdot 3^{-x} \cdot \ln 3 \cdot (-1) = -3 \cdot 3^{-x} \ln 3$ $3^{-x} dx = -\frac{du}{3 \ln 3}$

Changing the limits of integration: When $x = 0$, $u = 1 + 3 \cdot 3^0 = 1 + 3 = 4$. When $x = 2$, $u = 1 + 3 \cdot 3^{-2} = 1 + \frac{3}{9} = 1 + \frac{1}{3} = \frac{4}{3}$.

Substituting these into the integral: $I = \int_4^{4/3} \frac{1}{u} \left( -\frac{du}{3 \ln 3} \right)$ $I = -\frac{1}{3 \ln 3} \int_4^{4/3} \frac{1}{u} du$ $I = \frac{1}{3 \ln 3} \int_{4/3}^4 \frac{1}{u} du$ $I = \frac{1}{3 \ln 3} [\ln |u|]_{4/3}^4$ $I = \frac{1}{3 \ln 3} \left( \ln 4 - \ln \frac{4}{3} \right)$ $I = \frac{1}{3 \ln 3} \left( \ln 4 - (\ln 4 - \ln 3) \right)$ $I = \frac{1}{3 \ln 3} (\ln 3)$ $I = \frac{1}{3}$

Therefore, the value of the definite integral is $\frac{1}{3}$.

Hence, option (B) is correct.