\( \quad a \int_{-a}[\\{(|x+a|) /(x+a)\\}+\\{(|x-a|) /(x-a)\\}] d x\) is equal to \(\ldots \ldots\) (where \(a>0)\) (a) 0 (b) a (c) \(2 \mathrm{a}\) (d) \(4 \mathrm{a}\)

When integrating expressions involving absolute values, it's essential to understand that the absolute value function, denoted by \( |x| \), defines the distance of \(x\) from zero on the real number line, taking only non-negative values. To integrate an absolute value, you must consider the function as piecewise-defined—essentially breaking the integral down where the inside of the absolute value changes sign.

This requires splitting the integral at points where \(x\) makes the inside of the absolute value zero and then considering the positive and negative scenarios separately. For example, with \( |x+a| \), if \(x > -a\), \(|x + a|\) is \(x + a\). If \(x < -a\), it is \(-x - a\) because absolute values turn negative expressions positive. Subsequently, you integrate each piece according to the standard rules of integration and combine the results.

A piecewise function is a function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. These functions are typically represented by different formulas for different parts of their domain.

To integrate a piecewise function, you need to handle each segment individually over its specific interval, ensuring continuity at the borders between pieces. In the given exercise, the function inside the integrand is piecewise because of the absolute values, with different expressions evaluated on either side of \(-a\) and \(a\). It's like assembling a puzzle, fitting each integral piece exactly where it belongs along the x-axis, then calculating the total area under the curve.

The Fundamental Theorem of Calculus is a central principle that connects differentiation with integration, providing a method for evaluating definite integrals. It states that if a function \(f\) is continuous on the closed interval \([a, b]\) and \(F\) is an antiderivative of \(f\) on \([a, b]\), then:
\[\int_{a}^{b} f(x) dx = F(b) - F(a)\]
Essentially, this theorem lets you take the antiderivative of the function at the upper bound and subtract the antiderivative at the lower bound to find the definite integral. This is exactly what allows the calculation of the area under the curve for the second integral in the given exercise, moving from evaluating the integral to computing the actual numbers that represent the accumulated area.

Evaluating a definite integral involves finding the net area between the function graph and the x-axis, within specific limits. You'll use the Fundamental Theorem of Calculus, which tells us to take the difference between the values of an antiderivative evaluated at the upper and lower limits.

In our exercise, after recognizing the piecewise nature of the integrand, we're left with evaluating \(a\) times the integral from \(-a\) to \(a\) of \(2dx\), which is straightforward. The process of evaluation gives us a clear numeric answer, representing the area under the curve between \(-a\) and \(a\), which in this case turns out to be \(4a\) — matching choice (d).

Exercise Improvement Advice

To improve your understanding, visualize the piecewise function graph, and clearly delineate the intervals. Always reassess the original function's behavior at the interval endpoints to ensure continuity and correctly apply the Fundamental Theorem of Calculus.