Evaluate the definite integral from 1 to -3 of (-x/2+1) dx

Question

Definite Integrals

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푑푥



Answer 1

∫_{(1 to -3)} (-x/2 + 1)dx = -∫_{(-3 to 1)} (-x/2 + 1)dx = -[-(1/4)x² + x]_{(-3 to 1)}

= [(1/4)x² - x]_{(-3 to 1)} = [1/4 - 1] - [9/4 + 3] = -6