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Question : 145 of 150
Marks:
+1,
-0
Solution:
Let
I=log‌x‌dxHere, we see that
log‌x is negative for
x∈(1,‌).
∴‌‌I‌=−1×(log‌x)‌dx‌=−[log‌x×x−∫‌×x‌dx]1∕e1‌=−[x‌log‌x−x]1∕e1‌=−[1‌log‌1−1−(‌‌log‌−‌)] =−[0−1−{‌(log‌1−log‌e)}−‌]=−[−1−{‌(0−1)−‌}]=1+(−‌−‌)=1−‌
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