Multivariate Calculus 10B, Problem 1

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, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)}