5.3 The Fundamental Theorem of Calculus/9 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/9" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T20:14:54Z

Protected "5.3 The Fundamental Theorem of Calculus/9" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 20:14, 6 September 2022
(No difference)

2022-09-06T19:56:42Z

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2022-09-06T19:55:14Z

← Older revision		Revision as of 19:55, 6 September 2022
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	\frac{d}{dy}\left[g(y)\right] = \frac{d}{dy}\left[\int_{2}^{y}t^2\sin{(t)}dt\right]		\frac{d}{dy}\left[g(y)\right] = \frac{d}{dy}\left[\int_{2}^{y}t^2\sin{(t)}dt\right]

	= 1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})~~</math><br>~~		= 1\cdot((y)^{2}\sin{y})-0\cdot((0)^{2}\sin{(2)})
	~~<math>~~= y^{2} sin{(y)}</math>		= y^{2}\sin{(y)}

			</math>

2022-09-06T19:52:58Z

← Older revision		Revision as of 19:52, 6 September 2022
Line 5:		Line 5:
	\frac{d}{dy}\left[g(y)\right] = \frac{d}{dy}\left[\int_{2}^{y}t^2\sin{(t)}dt\right]		\frac{d}{dy}\left[g(y)\right] = \frac{d}{dy}\left[\int_{2}^{y}t^2\sin{(t)}dt\right]

	~~<math>~~= 1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})</math><br>		= 1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})</math><br>
	<math>= y^{2} sin{(y)}</math>		<math>= y^{2} sin{(y)}</math>

2022-09-06T19:52:43Z

← Older revision		Revision as of 19:52, 6 September 2022
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	<math>g(y)= \int_{2}^{y}t^2\sin{(t)}dt</math><br>		<math>g(y)= \int_{2}^{y}t^2\sin{(t)}dt</math><br><br>

			<math>

			\frac{d}{dy}\left[g(y)\right] = \frac{d}{dy}\left[\int_{2}^{y}t^2\sin{(t)}dt\right]

	<math>= 1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})</math><br>		<math>= 1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})</math><br>
	<math>= y^{2} sin{(y)}</math>		<math>= y^{2} sin{(y)}</math>

2022-08-26T18:47:16Z

← Older revision		Revision as of 18:47, 26 August 2022
Line 1:		Line 1:
	<math>g(y)= \int_{2}^{y}t^2\sin{(t)}dt</math><br>		<math>g(y)= \int_{2}^{y}t^2\sin{(t)}dt</math><br>
	<math>1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})=		<math>= 1\cdot(y^{2}sin{(y)})-0\cdot(2^{2}sin{(2)})</math><br>
	y^{2} sin{(y)}</math>		<math>= y^{2} sin{(y)}</math>