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Question

Mathematics Question on integral

∫13∣(2−x)log⁡x∣dx\int \limits_1^3|(2-x) \log x|dx is equal to

A

32log⁡ 3+12\frac{3}{2} \log \, 3 + \frac{1}{2}

B

log⁡1633−12\log \frac{16}{3 \sqrt{3}}- \frac{1}{2}

C

−32log⁡3−12-\frac{3}{2} \log 3 - \frac{1}{2}

D

none of these.

Answer

log⁡1633−12\log \frac{16}{3 \sqrt{3}}- \frac{1}{2}

Explanation

Solution

∫13∣(2−x)log x∣dx=∫12∣2−x∣∣log x∣dx\int\limits_{1}^{3} \left|\left(2-x\right)log\,x\right|dx=\int\limits_{1}^{2}\left|2-x\right|\left|log\,x\right|dx +∫23∣2−x∣∣log x∣dx+\int\limits_{2}^{3}\left|2-x\right|\left|log\,x\right|dx =∫12(2−x)log x dx+∫23(x−2)log x dx=\int\limits_{1}^{2}\left(2-x\right)log \, x\,dx+\int\limits_{2}^{3}\left(x-2\right)log\,x \, dx =∣log x(2x−x22)∣12−∫121x(2x−x22)dx=\left|log\,x\left(2x-\frac{x^{2}}{2}\right)\right|_{1}^{2}-\int_{1}^{2} \frac{1}{x} \left(2x-\frac{x^{2}}{2}\right)dx +∣log x(x22−2x)∣23−∫231x(x2−2x)dx+\left|log\,x\left(\frac{x^{2}}{2}-2x\right)\right|_{2}^{3}-\int_{2}^{3} \frac{1}{x}\left(x^{2}-2x\right)dx =4log 2−32log3−12=4 log\, 2-\frac{3}{2} log 3 -\frac{1}{2} =log1633−12=log \frac{16}{3\sqrt{3}}-\frac{1}{2}