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DEFINITE INTEGRALS WITH ABSOLUTE VALUE This note addresses the calculation of definite integrals of the form ˆ b |f (x)| dx a when f (x) is a linear or quadratic polynomial. The absolute value function ( x, 0x (1) |x| = x, x < 0 has the familiar “V”-shaped graph. Integrands featuring absolute value must be split up according to this piecewise definition. Following are two examples of this process. ´ 10 Example 1. Find 0 |x 5| dx. Solution. Replacing x with x 5 in equation (1) gives ( x 5, 5 x |x 5| = 5 x, x < 5 where we’ve rewritten 0 x 5 and x 5 < 0 as 5 x and x < 5. Since 5 is in the interval of integration [0, 10], we have to use both parts of this definition. (If we integrate over [6, 10] or [0, 4], for example, then we only need to use the first line of the definition.) Therefore ˆ 10 ˆ 5 ˆ 10 |x 5| dx = |x 5| dx + |x 5| dx 0 0 5 ˆ 5 ˆ 10 = 5 x dx + x 5 dx 0 5 10 1 25 1 = 5x x + x2 5x 2 0 2 5 ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ 1 2 1 2 1 1 2 2 = 5(5) (5) 5(0) (0) + (10) 5(10) (5) 5(5) 2 2 2 2 = 25 ⇤ using FTC2. The crucial steps in example 1 are to determine the values of x where the definition of |x 5| changes, determine whether they are in the interval of integration, then split the integral into pieces accordingly. The same steps apply in general. (1) Determine the values of x when the definition of |f (x)| changes. Say they are a < x1 < x2 < · · · < xn < b. (2) Check which (if any) of these values lie in the interval of integration [a, b]. 1 (3) Split the integral of |f (x)| over [a, b] into several pieces ˆ b ˆ x1 ˆ x2 ˆ (2) |f (x)| dx = |f (x)| dx + |f (x)| dx + · · · + a a x1 b xn |f (x)| dx. Replace each |f (x)| by either f (x) or f (x), depending on the details of your particular problem. This turns the problem of finding the integral of |f (x)| into the problems of finding integrals of f (x), f (x). In example 1 we had only x1 = 5, and so we ended up with two integrals. Example 2. Set up the equation (2) for the absolute value integral ˆ 5 | t2 + 6t 8| dt. 2 Solution. For ease of reference call that function v(t). To evaluate this integral we must understand the sign of v(t) = (t 4)(t 2) on the interval [ 2, 5]. Because the roots of v(t) are t = 2, 4 and the leading coeﬃcient of v(t) is negative, it follows that v(t) is negative on [ 2, 2], positive on [2, 4], and negative again on [4, 5]. You can graph this to see for yourself. This means the definition of |v(t)| changes at t1 = 2, t2 = 4, t3 = 5. Accordingly we must split up [ 2, 5] = [ 2, 2] [ [2, 4] [ [2, 5] and use the definitions according to equation (1) 8 > 2t2 < v(t), |v(t)| = v(t), 2t4 > : v(t), 4 t 5 Therefore ˆ 5 2 |v(t)| dt = ˆ = ˆ 2 v(t) dt + 2 2 4 v(t) dt + 2 t 2 ˆ 2 6t + 8 dt + ˆ 5 v(t) dt 4 ˆ 2 4 2 t + 6t 8 dt + ˆ 5 t2 6t + 8 dt 4 We leave this unevaluated for the sake of neatness, but the reader should be able to use FTC2 and linearity/power rule to calculate these three definite integrals. The final answer is 40. ⇤ 2