18 -S x+3 dx = lim [SI 9-3 -S x+3 b-3* = lim a-3 18 9 1₁-0 dx = x+3 -5 (If the integral does not exist, enter 'DNE'.) •Discontinuous at x = -3 9 + lim 18 lim b-3+ 9 x+3 () () a+3 (-5)+3 18+3 b+3 Use FTC 2 JUse

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