To find a function \( f(x) \) that satisfies the condition \(\int_{-2}^{2} f(x) \, dx = 0\), we must consider the properties of definite integrals. Specifically, this condition implies that the net area under the curve of \( f(x) \) from \( x = -2 \) to \( x = 2 \) is zero. This usually means that the function is symmetric about the y-axis (an odd function) or has equal positive and negative areas in this interval.
This function is an odd function because replacing \( x \) with \(-x\) gives:
\(\log\left(\frac{2-x}{2+x}\right) = -\log\left(\frac{2+x}{2-x}\right)\).Since it is an odd function and the integration is symmetric around the y-axis over a symmetric interval \([-2, 2]\), the integral evaluates to zero:
\(\int_{-2}^{2} \log\left(\frac{2+x}{2-x}\right) \, dx = 0\).The function \(\sin(2+x)\) is not symmetric about the y-axis in the interval \([-2, 2]\), and thus does not guarantee that the integral is zero:
\(\int_{-2}^{2} \sin(2+x) \, dx \neq 0\).This polynomial does not adhere to symmetry properties that ensure zero integral over the given interval. Evaluating:
\(\int_{-2}^{2} (2x^3 + 2x + 1) \, dx\)would require computing each term separately, and clearly, it does not yield zero.
The function \(2x \tan x\) does not have symmetry about the y-axis in the interval \([-2, 2]\) and hence does not guarantee the integral resolves to zero:
\(\int_{-2}^{2} 2x \tan x \, dx \neq 0\).Conclusion: The function \(\log\left(\frac{2+x}{2-x}\right)\) is the only function among the options provided that satisfies the condition \(\int_{-2}^{2} f(x) \, dx = 0\) due to its odd symmetry.