Answer to Question #3633 in Calculus for noor
What is the integrate of
∫ e [sup](x)^1/2[/sup] dx
1
2011-07-18T07:37:15-0400
<img src="/cgi-bin/mimetex.cgi?%5Cint%7Be%5E%7B%5Csqrt%7Bx%7D%7D%20dx%7D%20=%20[%5Csqrt%7Bx%7D%20=%20t]%20=%20%5Cint%7Be%5Et%20dt%5E2%7D%20=%202%20%5Cint%7Bt%20e%5Et%20dt%7D%20=%20t%20e%5Et%20-%20%5Cint%7B%20e%5Et%20dt%7D%20=%20te%5Et%20-%20e%5Et%20=%20e%5Et%20%28t-1%29%20=%20e%5E%7B%5Csqrt%7Bx%7D%7D%28%5Csqrt%7Bx%7D%20-1%29" title="\int{e^{\sqrt{x}} dx} = [\sqrt{x} = t] = \int{e^t dt^2} = 2 \int{t e^t dt} = t e^t - \int{ e^t dt} = te^t - e^t = e^t (t-1) = e^{\sqrt{x}}(\sqrt{x} -1)">
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