Answer:
C. 14
Explanation:
F(x) = ∫₀²ˣ √(t³−15) dt
Use second fundamental theorem of calculus:
F'(x) = √((2x)³−15) d/dx (2x)
F'(x) = 2 √(8x³−15)
Evaluate at x=2:
F'(2) = 2 √(8×2³−15)
F'(2) = 2 √(64−15)
F'(2) = 2 √49
F'(2) = 14
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