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Find the integral?

Find the integral of 2^5x dx from negative infinity to 0.

When I integrate, I got 2^(5x)/(5*ln(2)) from negative infinity to 0, and I got 1/(5*ln(2))+infinity, but the answer should be positive infinity, why is the answer in the book says 1/(5*ln(2))? What's the correct answer? Show work.


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Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
Check Marked as Best Answer
∫{-∞, 0}2^(5x) dx
= (1/5)∫{-∞, 0}2^(5x) d5x
= (1/5)2^(5x)/ln2 from -∞ to 0
= 1/(5ln2) <==Answer
Attn: (1/5)2^(5x)/ln2 at x = -∞ is equal to zero.
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
-∞0 25xdx= 1/5∫-∞02zdz=1/5(2z/ln(2))|-∞0=1/(5*ln(2))[20-limz→-∞2z]=1/(5*ln(2))[1-0]=1/(5*ln(2))
Here I made a change of variable z=5x. Lim 2z when z goes to - infinity is obviously zero.