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Answer to Question #3810 in Calculus for noor

Question #3810
Evaluate the improper integral &
1-& & & ʃ& & In x& / x& & dx& & & & & & & & & & & & & & & & & & & from (& & 0& & & to& & & 4 )
2-& & & ʃ& & dx& & /& & (& 4 – x )1/5& & & & & & & & & & & from& (& & 0& & to& & & & 4)
3-& & & ʃ& & & 1 / x2& & & & dx& & & & & & & & & & & & & & & & & & & & & from (& & -∞& & & & & & to& -3& & ) &
4-& & & ʃ& & 1/ x4& & & dx& & & & & & & & & & & & & & & & & & & & from (& & -∞& & & & & & to& -2& & )
Expert's answer
<img src="/cgi-bin/mimetex.cgi?%5Cint_0%5E4%20%5Cfrac%7B%5Cln%20x%7D%7Bx%7Ddx%20=[this%20%5C%20integral%20%5C%20does%20%5C%20not%20%5C%20converge]%20=%20PV%5Cint_%7B0%7D%5E%7B4%7D%20%5Cln%20x%20%5C%20d%7B%5Cln%20x%7D%20=%20-%20%5Cinfty%20%5C%5C%20%5Cint_0%5E4%20%5Cfrac%7Bdx%7D%7B%284-x%29%5E%7B1/5%7D%7D%20=%20-%20%5Cint_0%5E4%20%5Cfrac%7Bd%284-x%29%7D%7B%284-x%29%5E%7B1/5%7D%7D%20=%20-%20%5Cfrac%7B5%7D%7B4%7D%20%7B%284-x%29%5E%7B4/5%7D%7D%7C_0%5E4%20=%20%5Cfrac%7B5%7D%7B4%7D4%5E%7B4/5%7D%20=%20%5Cfrac%7B5%7D%7B4%5E%7B1/5%7D%7D%20%5C%5C%20%5Cint_%7B-%5Cinfty%7D%5E%7B-3%7D%20%5Cfrac%7Bdx%7D%7Bx%5E2%7D%20=%20-%20%5Cfrac%7B1%7D%7Bx%7D%20%7C_%7B-%5Cinfty%7D%5E%7B-3%7D%20=%20%5Cfrac%7B1%7D%7B3%7D%20-%20%5Cfrac%7B1%7D%7B-%20%5Cinfty%7D%20=%20%5Cfrac%7B1%7D%7B3%7D%20%5C%5C%20%5Cint_%7B-%5Cinfty%7D%5E%7B-2%7D%20%5Cfrac%7Bdx%7D%7Bx%5E4%7D%20=%20-%20%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B1%7D%7Bx%5E3%7D%20%7C_%7B-%5Cinfty%7D%5E%7B-2%7D%20=%20-%20%5Cfrac%7B1%7D%7B3*%28-2%29%5E3%7D%20=%20%5Cfrac%7B1%7D%7B24%7D" title="\int_0^4 \frac{\ln x}{x}dx =[this \ integral \ does \ not \ converge] = PV\int_{0}^{4} \ln x \ d{\ln x} = - \infty \\ \int_0^4 \frac{dx}{(4-x)^{1/5}} = - \int_0^4 \frac{d(4-x)}{(4-x)^{1/5}} = - \frac{5}{4} {(4-x)^{4/5}}|_0^4 = \frac{5}{4}4^{4/5} = \frac{5}{4^{1/5}} \\ \int_{-\infty}^{-3} \frac{dx}{x^2} = - \frac{1}{x} |_{-\infty}^{-3} = \frac{1}{3} - \frac{1}{- \infty} = \frac{1}{3} \\ \int_{-\infty}^{-2} \frac{dx}{x^4} = - \frac{1}{3} \frac{1}{x^3} |_{-\infty}^{-2} = - \frac{1}{3*(-2)^3} = \frac{1}{24}">

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