Difference between revisions of "Multivariate Calculus 10B, Problem 1"

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function draw() {
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  var canvas = document.getElementById('canvas');
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  if (canvas.getContext){
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    var ctx = canvas.getContext('2d');
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    ctx.beginPath();
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    ctx.moveTo(75,50);
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    ctx.lineTo(100,75);
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    ctx.lineTo(100,25);
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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>
 
|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>

Revision as of 22:29, 7 February 2016

Calculate the following integrals

a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int _0^1 \int_y^1 e^{\frac{y}{x}}~dxdy}
b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^1 \int_0^{cos^{-1}(y)} e^{2x-y}~dxdy}


solution(a):

Here we change order of integration, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)}

solution(b):

function draw() {

 var canvas = document.getElementById('canvas');
 if (canvas.getContext){
   var ctx = canvas.getContext('2d');
   ctx.beginPath();
   ctx.moveTo(75,50);
   ctx.lineTo(100,75);
   ctx.lineTo(100,25);
   
 }

}

Here we change order of integration, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)}