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The value of the integral `int_(e^(-1))^(e^2)|((log)_e x)/x|dxi s``3/2`(b) `5/2`(c) 3(d) 5A. `3//2`B. `5//2`C. 3D. 5 |
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Answer» Correct Answer - D `int_(e^(-1))^(e^(2))|(log_(e)x)/(x)|dx=int_(e^(-1))^(1)|(log_(e)x)/(x)|dx-int_(1)^(e^(2))|(log_(e)x)/(x)|dx` `["since , 1 is turning point for"|(log_(e)x)/(x)|" for + ve and - ve values"]` `=-int_(e^(-1))^(1)(log_(e)x)/(x)dx+int_(1)^(e^(2))|(log_(e)x)/(x)|dx` `=-(1)/(2)[(log_(e)x)^(2)]_(e^(-1))^(1)+(1)/(2)[(log_(e)x)^(2)]_(1)^(e^(2))` `=-(1)/(2){0-(-1)^(2)}+(1)/(2)(2^(2)-0)=(5)/(2)` |
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