∫\intop∫ 2-1 1/ 3√x2 dx=?
∫−1213x2dx=∫−1213∗∣x∣dx=∫−1013∗(−x)dx+∫0213∗xdx=−13∫−101xdx+13∫021xdx=−13∗ln∣x∣∣x=−1x=0+13∗ln∣x∣∣x=0x=2=−13∗limx→0−ln(−x)+13∗ln(1)+13∗ln(2)−13∗limx→0+ln(x){\int_{-1}^{2} {\frac 1 {3\sqrt{x^2}}} dx}={\int_{-1}^{2} {\frac 1 {3*|x|}} dx}={\int_{-1}^{0} {\frac 1 {3*(-x)}} dx}+{\int_{0}^{2} {\frac 1 {3*x}} dx}=-{\frac 1 3}{\int_{-1}^{0} {\frac 1 x} dx}+{\frac 1 3}{\int_{0}^{2} {\frac 1 x} dx}=-{\frac 1 3}*ln|x||_{x=-1}^{x=0}+{\frac 1 3}*ln|x||_{x=0}^{x=2}=-{\frac 1 3}*{\lim_{x\to{0^-}} ln(-x)}+{\frac 1 3}*ln(1)+{\frac 1 3}*ln(2)-{\frac 1 3}*{\lim_{x\to{0^+}} ln(x)}∫−123x21dx=∫−123∗∣x∣1dx=∫−103∗(−x)1dx+∫023∗x1dx=−31∫−10x1dx+31∫02x1dx=−31∗ln∣x∣∣x=−1x=0+31∗ln∣x∣∣x=0x=2=−31∗limx→0−ln(−x)+31∗ln(1)+31∗ln(2)−31∗limx→0+ln(x)
Since limx→0+lnx=−∞{\lim_{x\to0^+} lnx}=-\inftylimx→0+lnx=−∞, then this integral is divergent and equals ∞\infty∞
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