Discussion
\(\int_{0}^{2}\frac{x^{4}}{x^{4} + (2 - x)^{4}}dx =\)
Answer: A
The integral is 1.
Replace \(x\) by \(2 - x\), which does not change a definite integral over \([0, 2]\):
\[ I = \int_{0}^{2}\frac{(2 - x)^{4}}{(2 - x)^{4} + x^{4}}dx \]
Adding the two forms, the integrand becomes 1, so \(2I = \int_{0}^{2}1\,dx = 2\) and \(I = 1\). 2 is \(2I\), forgetting to halve, and the other values do not follow.
This 'king' property, \(\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a - x)dx\), always gives half the length of the interval for integrands of the form \(\frac{g(x)}{g(x) + g(a - x)}\).
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