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

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|Here we use change of variable, <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</math>
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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>
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'''solution(b):'''
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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>

Latest revision as of 23:30, 7 February 2016

Calculate the following integrals

a)
b)


solution(a):

Here we change order of integration,

solution(b):

Here we change order of integration,