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{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> |
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):
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