Property Of Definite Integrals

Question

Assertion: the integral from zero to two of the function defined as x for x below one and as two minus x otherwise can be split additively; evaluate that integral.

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Piecewise integrands require the additivity property of definite integrals, which states that an integral over an interval equals the sum of integrals over any partition of that interval. The natural place to split is the point where the function definition changes, giving \int_0^2 f = \int_0^1 f + \int_1^2 f. On the first piece [0,1] we have f(x) = x, so \int_0^1 x\,dx = [x^2/2]_0^1 = 1/2. On the second piece [1,2] we have f(x) = 2 - x, so \int_1^2 (2 - x)\,dx = [2x - x^2/2]_1^2 = (4 - 2) - (2 - 1/2) = 2 - 3/2 = 1/2. Adding the two contributions gives 1/2 + 1/2 = 1. Option 2 either double-counts or mishandles the descending branch. Option 1/2 stops after evaluating only one piece. Option 3/2 mishandles the limits of the second integral. This additivity property is indispensable for piecewise and absolute-value integrands. As a final plausibility check, the graph forms a symmetric tent triangle with base 2 and height 1, whose area is \frac{1}{2}\cdot 2\cdot 1 = 1, matching the computed integral exactly and confirming geometric consistency.

Property Of Definite Integrals

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