Engineering Board Exam Reviewer · Definite Integrals · medium

Evaluate: -13≤ft|x2-x-2 |dx

1. 8/3
2. 19/3
3. 17/3
4. 11/3

Answer

B. 19/3

Solution

Factor x2-x-2=(x-2)(x+1), which is negative on (-1,2) and positive on (2,3). With (x2-x-2),dx=x3/3-x2/2-2x: ≤ft[-≤ft(x3/3-x2/2-2x ) ]-12+≤ft[x3/3-x2/2-2x ]23=9/2+11/6=19/3.

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