5.5 The Substitution Rule/63 - Revision history

Jorgen70034@students.laalliance.org at 00:59, 2 September 2022

2022-09-02T00:59:32Z

@@ Line 1: / Line 1: @@
-<math>\int_{0}^{a} x* \sqrt{x^2+a^2}dx</math>
+<math>\int_{0}^{a} x\,\cdot \sqrt{x^2+a^2}dx</math>
 <math>
@@ Line 16: / Line 16: @@
 \frac{1}{2}
 \int_{a^2}^{2a^2}
-\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot  \frac{a^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg|_{2a^2}^{a^2}
+\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot  \frac{a^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg|_{a^2}^{2a^2}
 \end{align}
 </math>

2022-09-01T20:23:38Z

← Older revision		Revision as of 20:23, 1 September 2022
Line 16:		Line 16:
	\frac{1}{2}		\frac{1}{2}
	\int_{a^2}^{2a^2}		\int_{a^2}^{2a^2}
	\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \frac{a^{\frac{1}{2}+1}}{1/2+1}\bigg\|_{2a^2}^{a^2}		\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \frac{a^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg\|_{2a^2}^{a^2}

	\end{align}		\end{align}
	</math>		</math>

2022-09-01T20:23:20Z

← Older revision		Revision as of 20:23, 1 September 2022
Line 16:		Line 16:
	\frac{1}{2}		\frac{1}{2}
	\int_{a^2}^{2a^2}		\int_{a^2}^{2a^2}
	\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \bigg\|_{2a^2}^{a^2}		\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \frac{a^{\frac{1}{2}+1}}{1/2+1}\bigg\|_{2a^2}^{a^2}

	\end{align}		\end{align}
	</math>		</math>

	~~\frac{a^{\frac{1}{2}+1}{1/2+1}~~

2022-09-01T20:22:55Z

← Older revision		Revision as of 20:22, 1 September 2022
Line 16:		Line 16:
	\frac{1}{2}		\frac{1}{2}
	\int_{a^2}^{2a^2}		\int_{a^2}^{2a^2}
	\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot ~~\frac{a^{\frac{1}{2}+1}{1/2+1}~~ \bigg\|_{2a^2}^{a^2}		\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \bigg\|_{2a^2}^{a^2}

	\end{align}		\end{align}
	</math>		</math>

			\frac{a^{\frac{1}{2}+1}{1/2+1}

2022-09-01T20:22:14Z

← Older revision		Revision as of 20:22, 1 September 2022
Line 17:		Line 17:
	\int_{a^2}^{2a^2}		\int_{a^2}^{2a^2}
	\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \frac{a^{\frac{1}{2}+1}{1/2+1} \bigg\|_{2a^2}^{a^2}		\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \frac{a^{\frac{1}{2}+1}{1/2+1} \bigg\|_{2a^2}^{a^2}
	{

	\end{align}		\end{align}
	</math>		</math>

2022-09-01T20:21:50Z

← Older revision		Revision as of 20:21, 1 September 2022
Line 16:		Line 16:
	\frac{1}{2}		\frac{1}{2}
	\int_{a^2}^{2a^2}		\int_{a^2}^{2a^2}
	\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \frac{a^{\frac{1/2}+1}{1/2+1} \bigg\|_{2a^2}^{a^2}		\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \frac{a^{\frac{1}{2}+1}{1/2+1} \bigg\|_{2a^2}^{a^2}
			{

	\end{align}		\end{align}
	</math>		</math>

2022-09-01T20:21:29Z

← Older revision		Revision as of 20:21, 1 September 2022
Line 16:		Line 16:
	\frac{1}{2}		\frac{1}{2}
	\int_{a^2}^{2a^2}		\int_{a^2}^{2a^2}
	\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot		\sqrt{u}\,\cdot du = \frac{1}{2}\,\cdot \frac{a^{\frac{1/2}+1}{1/2+1} \bigg\|_{2a^2}^{a^2}


	\end{align}		\end{align}
	</math>		</math>

2022-09-01T20:15:20Z

2022-09-01T19:52:06Z

2022-09-01T19:51:42Z