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MTH 1125 Calculus I Practice Quiz Exercises

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MTH 1125 W8FINAL EXAM 1. ‫׬‬ 𝑥 2 +5𝑥 𝑑𝑥 𝑥+7 ∫ Substitution 𝑥 2 + 5𝑥 𝑥2 𝑥 𝑑𝑥 = ∫ 𝑑𝑥 + 5 ∫ 𝑑𝑥 𝑥+7 𝑥+7 𝑥+7 𝐼: ∫ 𝑥2 𝑑𝑥 𝑥+7 Use substitution: 𝐿𝑒𝑡 𝑢 = 𝑥 + 7 → 𝑑𝑢 = {𝑑𝑥 2 1 𝑥 = (𝑢 − 7)2 = 𝑢2 − 14𝑢 + 49 𝑥2 𝑢2 − 14𝑢 + 49 𝑢2 𝑢 1 →∫ 𝑑𝑥 = ∫ 𝑑𝑢 = ∫ 𝑑𝑢 − 14 ∫ 𝑑𝑢 + 49 ∫ 𝑑𝑢 𝑥+7 𝑢 𝑢 𝑢 𝑢 = 𝑢2 − 14𝑢 + 49 ln|𝑢| 2 Replacing back 𝑢 = 𝑥 + 7: ∫ (𝑥 + 7)2 𝑥2 𝑑𝑥 = − 14(𝑥 + 7) + 49 ln|𝑥 + 7| 𝑥+7 2 𝐼𝐼: ∫ 𝑥 𝑑𝑥 𝑥+7 𝑑𝑢 Use substitution: 𝐿𝑒𝑡 𝑢 = 𝑥 + 7 → {𝑑𝑥 = 1 𝑥 =𝑢−7 →∫ 𝑥 𝑢−7 𝑢 1 𝑑𝑥 = ∫ 𝑑𝑢 = ∫ 𝑑𝑢 − 7 ∫ 𝑑𝑢 = 𝑢 − 7 ln|𝑢| 𝑥+7 𝑢 𝑢 𝑢 Replacing back 𝑢 = 𝑥 + 7: ∫ 𝑥 𝑑𝑥 = (𝑥 + 7) − 7 ln|𝑥 + 7| 𝑥+7 ⇒∫ ∴∫ (𝑥 + 7)2 𝑥2 𝑑𝑥 = − 14(𝑥 + 7) + 49 ln|𝑥 + 7| + 5((𝑥 + 7) − 7 ln|𝑥 + 7|) 𝑥+7 2 (𝑥 + 7)2 𝑥2 𝑑𝑥 = − 9𝑥 − 63 + 14 ln|𝑥 + 7| + 𝐶 𝑥+7 2 MTH 1125 W8FINAL EXAM 53 2. ‫׬‬1 5 𝑥 11 − 𝑥 2 𝑑𝑥 Fundamental Theorem 5 5 5 3 11 1 𝑥 −2+1 5 ∫ ( − 2 ) 𝑑𝑥 = 3 ∫ 𝑑𝑥 − 11 ∫ 𝑥 −2 𝑑𝑥 = 3 ln|𝑥| | − 11 ( )| 𝑥 𝑥 𝑥 −2 + 1 1 1 1 1 1 1 1 44 = 3[ln 5 − ln 1] + 11 [ − ] = 3 ln 5 − 5 1 5 ...
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