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Question Number 188861 by Shlock last updated on 08/Mar/23

Answered by cortano12 last updated on 08/Mar/23

$$\underset{0}{\stackrel{4}{\int }}\mathrm{x}\mathrm{d}\left(\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)\right)={[\mathrm{x}\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)]}_0^4$$$$=[4\mathrm{f}{}^{\prime }\left(4\right)-\mathrm{f}\left(4\right)+\mathrm{f}\left(0\right)]$$$$=4.\frac{1}{2}-2+3=3$$

Answered by mr W last updated on 08/Mar/23

$${\underset{0}{\int }}^4{xf}^{″}\left(x\right)dx$$$$={\underset{0}{\int }}^4{xdf}^{\prime }\left(x\right)$$$$={\left[{xf}^{\prime }\left(x\right)\right]}_0^4-{\underset{0}{\int }}^4{f}^{\prime }\left(x\right)dx$$$$={\left[{xf}^{\prime }\left(x\right)\right]}_0^4-{\left[f\left(x\right)\right]}_0^4$$$$=4f^{\prime }\left(4\right)-f\left(4\right)+f\left(0\right)$$$$=4\times \frac{2}{4}-2+3$$$$=3$$

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