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, $\int _{0}^{1}\int _{0}^{x}e^{\frac {y}{x}}~dydx=\int _{0}^{1}[xe^{\frac {y}{x}}|_{y=0}^{y=x}]~dx=\int _{0}^{1}x(e-1)~dx={\frac {1}{2}}x^{2}|_{0}^{1}(e-1)={\frac {1}{2}}(e-1)$ solution(b): function draw() { var canvas = document.getElementById('canvas'); if (canvas.getContext){ var ctx = canvas.getContext('2d'); ctx.beginPath(); ctx.moveTo(75,50); ctx.lineTo(100,75); ctx.lineTo(100,25); } } Here we change order of integration, $\int _{0}^{\frac {\pi }{2}}\int _{0}^{cos(x)}e^{2x-y}~dydx=\int _{0}^{\frac {\pi }{2}}[-e^{2x-y}|_{y=0}^{y=cos(x)}]~dx=\int _{0}^{\frac {\pi }{2}}[e^{2x}-e^{2x-cos(x)}]~dx=[{\frac {1}{2}}e^{2x}-{\frac {1}{2+sin(x)}}e^{2x-cos(x)}]|_{0}^{\frac {\pi }{2}}={\frac {e^{\pi }}{6}}+{\frac {1}{2}}({\frac {1}{e}}-1)$