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

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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>
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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>

Revision as of 23:11, 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,