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

Latest revision as of 23:30, 7 February 2016

Calculate the following integrals

a)

\int _{0}^{1}\int _{y}^{1}e^{\frac {y}{x}}~dxdy

b)

\int _{0}^{1}\int _{0}^{cos^{-1}(y)}e^{2x-y}~dxdy

solution(a):

Here we change order of integration,

\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):


Here we change order of integration, $\int _{0}^{\frac {\pi }{2}}\int _{0}^{cos(x)}e^{2x-y}~dydx=\int _{0}^{\frac {\pi }{2}}[-e^{2x-y}\|_{y=0}^{y=cos(x)}]~dx=\int _{0}^{\frac {\pi }{2}}[e^{2x}-e^{2x-cos(x)}]~dx=[{\frac {1}{2}}e^{2x}-{\frac {1}{2+sin(x)}}e^{2x-cos(x)}]\|_{0}^{\frac {\pi }{2}}={\frac {e^{\pi }}{6}}+{\frac {1}{2}}({\frac {1}{e}}-1)$

@@ Line 3: / Line 3: @@
 :: <span class="exam">b) <math>\int_0^1 \int_0^{cos^{-1}(y)} e^{2x-y}~dxdy</math>
-'''solution:'''
-'''a'''
+'''solution(a):'''
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+!
+|-
+|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>
- <math>\draw [thick] (-2,2) % Draws a line
+'''solution(b):'''
-      to [out=10,in=190] (2,2)
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-      to [out=10,in=90] (6,0)
+!
-      to [out=-90,in=30] (-2,-2);</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>