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Question : 77 of 160
Marks:
+1,
-0
Solution:
Consider the integral
I=e2|‌|dx It can be simplified as
I=‌1‌dx+‌e2‌dx Let
log‌x‌‌=t ‌‌‌=dt Thus,
I‌‌=tdt+tdt ‌‌=[‌]−10+[‌]02 ‌‌=−‌(0−1)+‌(4−0) ‌‌=‌
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