Multivariate Calculus 10B, Problem 1 - Revision history

James Ogaja 2 at 06:30, 8 February 2016

2016-02-08T06:30:58Z

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 |Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^{\frac{\pi}{2}} [e^{2x} - e^{2x - cos(x)}]~dx = [\frac{1}{2}e^{2x} - \frac{1}{2 + sin(x)}e^{2x - cos(x)}]|_0^{\frac{\pi}{2}} = \frac{e^{\pi}}{6} + \frac{1}{2}(\frac{1}{e} - 1)</math>

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+!
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 |Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^{\frac{\pi}{2}} [e^{2x} - e^{2x - cos(x)}]~dx = [\frac{1}{2}e^{2x} - \frac{1}{2 + sin(x)}e^{2x - cos(x)}]|_0^{\frac{\pi}{2}} = \frac{e^{\pi}}{6} + \frac{1}{2}(\frac{1}{e} - 1)</math>

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-|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^{\frac{\pi}{2}} [e^{2x} - e^{2x - cos(x)}]~dx = \frac{1}{2}e^{2x} - \frac{1}{2 + sin(x)}e^{2x - cos(x)}|_0^{\frac{\pi}{2}} = \frac{1}{2}(e - 1)</math>
+|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^{\frac{\pi}{2}} [e^{2x} - e^{2x - cos(x)}]~dx = [\frac{1}{2}e^{2x} - \frac{1}{2 + sin(x)}e^{2x - cos(x)}]|_0^{\frac{\pi}{2}} = \frac{e^{\pi}}{6} + \frac{1}{2}(\frac{1}{e} - 1)</math>

@@ Line 14: / Line 14: @@
+!
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-|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^{\frac{\pi}{2}} [e^{2x} - e^{2x - cos(x)}]~dx = \frac{1}{2}e^{2x} - \frac{1}{2 + sin(x)}e^{2x - cos(x)}|_0^{\frac{\pi}{2} = \frac{1}{2}(e - 1)</math>
+|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^{\frac{\pi}{2}} [e^{2x} - e^{2x - cos(x)}]~dx = \frac{1}{2}e^{2x} - \frac{1}{2 + sin(x)}e^{2x - cos(x)}|_0^{\frac{\pi}{2}} = \frac{1}{2}(e - 1)</math>

@@ Line 14: / Line 14: @@
+!
 |-
-|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^{\frac{\pi}{2}} [e^{2x} - e^{2x - cos(x)}]~dx = \frac{1}{2}x^2|_0^1(e - 1) = \frac{1}{2}(e - 1)</math>
+|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^{\frac{\pi}{2}} [e^{2x} - e^{2x - cos(x)}]~dx = \frac{1}{2}e^{2x} - \frac{1}{2 + sin(x)}e^{2x - cos(x)}|_0^{\frac{\pi}{2} = \frac{1}{2}(e - 1)</math>

@@ Line 14: / Line 14: @@
+!
 |-
-|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^1 x(e - 1)~dx = \frac{1}{2}x^2|_0^1(e - 1) = \frac{1}{2}(e - 1)</math>
+|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^{\frac{\pi}{2} [e^{2x} - e^{2x - cos(x)}]~dx = \frac{1}{2}x^2|_0^1(e - 1) = \frac{1}{2}(e - 1)</math>

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+!
 |-
-|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^1[xe^{\frac{y}{x}}|_{y = 0}^{y = x}]~dx = \int_0^1 x(e - 1)~dx = \frac{1}{2}x^2|_0^1(e - 1) = \frac{1}{2}(e - 1)</math>
+|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^{\frac{\pi}{2}[-e^{2x -y}|_{y = 0}^{y = cos(x)}]~dx = \int_0^1 x(e - 1)~dx = \frac{1}{2}x^2|_0^1(e - 1) = \frac{1}{2}(e - 1)</math>

← Older revision		Revision as of 05:55, 8 February 2016
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	\|Here we change order of integration, <math>\int _0^1 \int_0^x e^{\frac{y}{x}}~dydx = \int _0^1[xe^{\frac{y}{x}}\|_{y = 0}^{y = x}]~dx = \int_0^1 x(e - 1)~dx = \frac{1}{2}x^2\|_0^1(e - 1) = \frac{1}{2}(e - 1)</math>		\|Here we change order of integration, <math>\int _0^1 \int_0^x e^{\frac{y}{x}}~dydx = \int _0^1[xe^{\frac{y}{x}}\|_{y = 0}^{y = x}]~dx = \int_0^1 x(e - 1)~dx = \frac{1}{2}x^2\|_0^1(e - 1) = \frac{1}{2}(e - 1)</math>
		+
		+	'''solution(b):'''
		+	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"
		+	!
		+	\|-
		+	\|Here we change order of integration, <math>\int _0^{\frac{\pi}{2}} \int_0^{cos(x)} e^{2x - y}~dydx = \int _0^1[xe^{\frac{y}{x}}\|_{y = 0}^{y = x}]~dx = \int_0^1 x(e - 1)~dx = \frac{1}{2}x^2\|_0^1(e - 1) = \frac{1}{2}(e - 1)</math>