Step 1To find the definite integralThen the width of each interval isb-aAx =n6 - (-6)6+6Note that ||A||-n==Step 2Choose c; as the right endpoint of each subinterval. ThenC₁= a + i(Ax) =So the definite integral is given byS'oxidx -n→121.x³dx by the limit definition, divide the interval [-6, 6] inton subintervals.0 as n→ ∞0.limn→ ∞olim Σ f(c₁) Axi||4||→ 07 = 1limn → ∞0n→nlimn → ∞⁰n →limn→ ∞lim7 = 1Σ(1Σ|1lim₂3n-6 ++i-216 +-2592 +limi = 1nΣ f(c₁) Axii = 112/¹2²) (¹2²)n+;2+2 +nn-2592 +20736 nnªnn7 = 1n((₁+²)--)(+²)+n²-171/2) +n/ = 1in31 +;3nn++7)]

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To find the definite integral x³dx by the limit definition, divide the interval [-6, 6] into n subintervals. Then the width of each interval is b-a Ax = = 6 - (-6) 6+6 Note that ||A|| → 0 x³dx = n = = n Step 2 Choose c; as the right endpoint of each subinterval. Then C₁ = a + i(Ax) So the definite integral is given by = n→ 12 0 as n→ ∞0. n lim Σ f(c₁) Ax; ||A||→ 07 i = 1 lim n→ ∞ n n→ lim n→ ∞o = lim n→∞o n→ lim n→∞0 n -£(- i=1 lim i = 1 i = 1 -6 + lim + -216 + -2592 + lim + 12) (1²2) / = 1 n Σf(c) Axi i = 1 + ;2+ 2+ n n -2592 + 20736 n n n n² £³] 7 = 1 ](¹+ ²¹² ) 1 ¹¹) (¹²) n Σ i=1 + 17/1/2) + n³ 1 + n n ++/-)]