A function is continuous on and satisfies where for

MATHEMATICS · Single correct · AP · Calculus AB · Integration and Accumulation of Change · Integration and Accumulation of Change

Problem

A function hh is continuous on [0,7][0,7] and satisfies ∫03h(t) dt=5and∫37∣h(t)∣ dt=6,\displaystyle \int_{0}^{3} h(t)\,dt = 5 \quad\text{and}\quad \int_{3}^{7} |h(t)|\,dt = 6, where h(t)<0h(t)<0 for 3≤t≤73\le t\le 7. What is the value of ∫07h(t) dt\displaystyle \int_{0}^{7} h(t)\,dt? (a) 1111 (b) −11-11 (c) −1-1 (d) 11

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