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_0^1 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> |
Revision as of 22:58, 7 February 2016
Calculate the following integrals
- a)
- b)
solution(a):
Here we change order of integration,
solution(b):
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