Question

The value of the definite integral \( \int_0^2 \frac{1}{3^x + 3} dx \) is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{\log_e 3}{3}$
• D. $\frac{\log_e 3}{2}$

Hint:

For integrals of the form $\int \frac{1}{a^x + C} dx$, a common trick is to divide numerator and denominator by $a^x$ or use the symmetry property mentioned above. These usually lead to very simple results.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to evaluate a definite integral involving an exponential function in the denominator.

Step 2: Key Formula or Approach:

• Use the substitution $u = x - 1$ to centralize the integral around zero.

• Apply the property $\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$.

Step 3: Detailed Explanation:

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

• Factor out 3 from the denominator: $I = \int_0^2 \frac{1}{3(3^{x-1} + 1)} dx = \frac{1}{3} \int_0^2 \frac{1}{3^{x-1} + 1} dx$.

• Substitute $u = x - 1$, then $du = dx$. Limits: $x=0 \to u=-1$ and $x=2 \to u=1$.
\[ I = \frac{1}{3} \int_{-1}^1 \frac{1}{3^u + 1} du \]
• Apply the property: $\int_{-a}^a f(u) du = \int_0^a (f(u) + f(-u)) du$: \[ f(u) + f(-u) = \frac{1}{3^u + 1} + \frac{1}{3^{-u} + 1} = \frac{1}{3^u + 1} + \frac{3^u}{1 + 3^u} = \frac{1 + 3^u}{1 + 3^u} = 1 \]
• Thus, $I = \frac{1}{3} \int_0^1 1 du = \frac{1}{3} [u]_0^1 = \frac{1}{3}(1 - 0) = \frac{1}{3}$.

Step 4: Final Answer:
The value of the integral is $\frac{1}{3}$.