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# Answer to Question #3787 in Calculus for zizi

Question #3787
evalute the improper integral 1- In x / x dx E( 0 , 4 ) 2- x e^-x^2 dx E( - infinity to + infinity)
<img src="/cgi-bin/mimetex.cgi?%5Cint_0%5E4%20%5Cfrac%7B%5Cln%7Bx%7D%20dx%7D%7Bx%7D%20=%20%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cln4%7D%20%5Cln%7Bx%7D%20d%20%5Cln%7Bx%7D%20=%20%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cln4%7D%20t%20d%20t%20=%20t%5E2%20%7C_%7B-%5Cinfty%7D%5E%7B%5Cln4%7D%20=%20%28%5Cln4%29%5E2%20-%20%28-%5Cinfty%29%5E2%20=%20-%5Cinfty%20%5C%5C%20%5Cint_%7B-%5Cinfty%7D%5E%7B+%5Cinfty%7D%20x%20e%5E%7B-x%5E2%7D%20dx%20=%20%5Cint_%7B+%5Cinfty%7D%5E%7B+%5Cinfty%7D%201/2%5C%20e%5E%7B-x%5E2%7D%20dx%5E2%20=%200" title="\int_0^4 \frac{\ln{x} dx}{x} = \int_{-\infty}^{\ln4} \ln{x} d \ln{x} = \int_{-\infty}^{\ln4} t d t = t^2 |_{-\infty}^{\ln4} = (\ln4)^2 - (-\infty)^2 = -\infty \\ \int_{-\infty}^{+\infty} x e^{-x^2} dx = \int_{+\infty}^{+\infty} 1/2\ e^{-x^2} dx^2 = 0">

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