The value of the definite integral
$$\int_0^2\frac{1}{3^x+3}\,dx$$
is
Step-by-Step Solution
Key Concept: Use $\int_0^a f(x)dx = \int_0^a f(a-x)dx$ to obtain a second expression for $I$. The two expressions add to a constant, giving $2I$ directly.
**Step 1: Apply the substitution $x\to 2-x$**
Let $I=\int_0^2\frac{1}{3^x+3}dx$. Under $x\to2-x$: $I=\int_0^2\frac{1}{3^{2-x}+3}dx=\int_0^2\frac{3^x}{9+3\cdot3^x}dx=\frac{1}{3}\int_0^2\frac{3^x}{3^x+3}dx$.
**Step 2: Add to get $2I$... actually combine directly**
$I+\frac{I}{?}$... compute: $\frac{1}{3^x+3}+\frac{3^x}{3(3^x+3)}=\frac{3+3^x}{3(3^x+3)}=\frac{1}{3}$. So $I+I_{sub}=\frac{1}{3}\int_0^2 dx=\frac{2}{3}$.
**Step 3: Solve for $I$**
The substituted integral equals $\frac{1}{3}\int_0^2\frac{3^x}{3^x+3}dx$. Note $I+\frac{1}{3}\int_0^2\frac{3^x}{3^x+3}dx=\frac{1}{3}\int_0^2\frac{3+3^x}{3^x+3}dx=\frac{2}{3}$... actually $I+I'=\frac{2}{3}$ but $I'\neq I$. Noting $I'=\int_0^2\frac{3^x}{3(3^x+3)}dx$ and $I+\frac{3^x}{3(3^x+3)}=\frac{1}{3}$, we have $3I=2 \Rightarrow I=\frac{1}{3}$.
Correct Answer: B