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MTH 1125 Calculus I Practice Quiz Exercises

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MTH 1125 W8FINAL EXAM 1. ‫׬‬ π‘₯ 2 +5π‘₯ 𝑑π‘₯ π‘₯+7 ∫ Substitution π‘₯ 2 + 5π‘₯ π‘₯2 π‘₯ 𝑑π‘₯ = ∫ 𝑑π‘₯ + 5 ∫ 𝑑π‘₯ π‘₯+7 π‘₯+7 π‘₯+7 𝐼: ∫ π‘₯2 𝑑π‘₯ π‘₯+7 Use substitution: 𝐿𝑒𝑑 𝑒 = π‘₯ + 7 β†’ 𝑑𝑒 = {𝑑π‘₯ 2 1 π‘₯ = (𝑒 βˆ’ 7)2 = 𝑒2 βˆ’ 14𝑒 + 49 π‘₯2 𝑒2 βˆ’ 14𝑒 + 49 𝑒2 𝑒 1 β†’βˆ« 𝑑π‘₯ = ∫ 𝑑𝑒 = ∫ 𝑑𝑒 βˆ’ 14 ∫ 𝑑𝑒 + 49 ∫ 𝑑𝑒 π‘₯+7 𝑒 𝑒 𝑒 𝑒 = 𝑒2 βˆ’ 14𝑒 + 49 ln|𝑒| 2 Replacing back 𝑒 = π‘₯ + 7: ∫ (π‘₯ + 7)2 π‘₯2 𝑑π‘₯ = βˆ’ 14(π‘₯ + 7) + 49 ln|π‘₯ + 7| π‘₯+7 2 𝐼𝐼: ∫ π‘₯ 𝑑π‘₯ π‘₯+7 𝑑𝑒 Use substitution: 𝐿𝑒𝑑 𝑒 = π‘₯ + 7 β†’ {𝑑π‘₯ = 1 π‘₯ =π‘’βˆ’7 β†’βˆ« π‘₯ π‘’βˆ’7 𝑒 1 𝑑π‘₯ = ∫ 𝑑𝑒 = ∫ 𝑑𝑒 βˆ’ 7 ∫ 𝑑𝑒 = 𝑒 βˆ’ 7 ln|𝑒| π‘₯+7 𝑒 𝑒 𝑒 Replacing back 𝑒 = π‘₯ + 7: ∫ π‘₯ 𝑑π‘₯ = (π‘₯ + 7) βˆ’ 7 ln|π‘₯ + 7| π‘₯+7 β‡’βˆ« ∴∫ (π‘₯ + 7)2 π‘₯2 𝑑π‘₯ = βˆ’ 14(π‘₯ + 7) + 49 ln|π‘₯ + 7| + 5((π‘₯ + 7) βˆ’ 7 ln|π‘₯ + 7|) π‘₯+7 2 (π‘₯ + 7)2 π‘₯2 𝑑π‘₯ = βˆ’ 9π‘₯ βˆ’ 63 + 14 ln|π‘₯ + 7| + 𝐢 π‘₯+7 2 MTH 1125 W8FINAL EXAM 53 2. ‫׬‬1 5 π‘₯ 11 βˆ’ π‘₯ 2 𝑑π‘₯ Fundamental Theorem 5 5 5 3 11 1 π‘₯ βˆ’2+1 5 ∫ ( βˆ’ 2 ) 𝑑π‘₯ = 3 ∫ 𝑑π‘₯ βˆ’ 11 ∫ π‘₯ βˆ’2 𝑑π‘₯ = 3 ln|π‘₯| | βˆ’ 11 ( )| π‘₯ π‘₯ π‘₯ βˆ’2 + 1 1 1 1 1 1 1 1 44 = 3[ln 5 βˆ’ ln 1] + 11 [ βˆ’ ] = 3 ln 5 βˆ’ 5 1 5 ...
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