User contributions for Dvaezazizi@laalliance.org

26 August 2022

• 20:2020:20, 26 August 2022 diff hist −894‎ 6.1 Areas Between Curves ‎ →‎Solutions
• 20:1820:18, 26 August 2022 diff hist +1,386‎ N 6.1 Areas Between Curves ‎ Created page with "== Lecture == == Lecture notes == == Solutions == Mr. V solutions: 2, 30, 54<br> === Block #1 === 1 2 3 5 === Block #2 === 7 9 11 13 15 17 5.5 T..."
• 20:1620:16, 26 August 2022 diff hist +96‎ Calculus J. Stewart - 6th Edition ‎ →‎Applications of Integrals
• 19:3119:31, 26 August 2022 diff hist +5‎ 5.5 The Substitution Rule/54 ‎No edit summary
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• 19:1819:18, 26 August 2022 diff hist 0‎ 5.5 The Substitution Rule ‎ →‎Solutions
• 19:1819:18, 26 August 2022 diff hist 0‎ 5.5 The Substitution Rule ‎ →‎Block #3
• 19:1819:18, 26 August 2022 diff hist +42‎ N 5.5 The Substitution Rule/52 ‎ Dvaezazizi@laalliance.org moved page 5.5 The Substitution Rule/52 to 5.5 The Substitution Rule/54 current Tag: New redirect
• 19:1819:18, 26 August 2022 diff hist 0‎ m 5.5 The Substitution Rule/54 ‎ Dvaezazizi@laalliance.org moved page 5.5 The Substitution Rule/52 to 5.5 The Substitution Rule/54
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• 19:1519:15, 26 August 2022 diff hist +405‎ N 5.5 The Substitution Rule/54 ‎ Created page with "<math> \int_{0}^{7} \sqrt{4+3x}dx </math> <math> \begin{align} u &=\ln(x) \\[2ex] du &= \frac{1}{x}dx \\[2ex] \end{align} </math> <math> \begin{align} \int \frac{\sin{(\ln{(x))}}}{x}dx &= \int\frac{1}{x}\sin(\ln{(x)})dx = \int\left(\frac{1}{x}dx\right)\sin{(\ln{(x)})} \\[2ex] &= \int (du)\sin{(u)} = \int \sin{(u)}du \\[2ex] &= -\cos{(u)} + C \\[2ex] &= -\cos{(\ln{(x)})} + C \end{align} </math>"
• 19:1019:10, 26 August 2022 diff hist 0‎ m 5.5 The Substitution Rule/30 ‎ Protected "5.5 The Substitution Rule/30" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite)) current
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• 19:0219:02, 26 August 2022 diff hist +419‎ N 5.5 The Substitution Rule/30 ‎ Created page with "<math> \int \frac{\sin{\ln{(x)}}}{x}dx </math> <math> \begin{align} u &=2+x^4 \\[2ex] du &= 4x^3dx \\[2ex] \frac{1}{4}du &= x^3dx \end{align} </math> <math> \begin{align} \int x^3(2+x^4)^5dx &= \int (x^3dx)(2+x^4) = \int \left(\frac{1}{4}du\right)(u) = \frac{1}{4}\int u\,du \\[2ex] &= \frac{1}{4} \left[\frac{u^{1+1}}{1+1}\right] + C = \frac{u^2}{8} + C \\[2ex] &= \frac{(2+x^4)^2}{8} + C \end{align} </math>"
• 19:0019:00, 26 August 2022 diff hist 0‎ m 5.5 The Substitution Rule/2 ‎ Protected "5.5 The Substitution Rule/2" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite)) current
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