lilybeth647321
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Hi im having a problem with this question:
(iii) Show that the integral of 1/(4x^2 -1)^0.5 (with lower bound 0.625 and upper bound 1.3) = 0.5(ln(5)-ln(2))

I took out a multiple of 1/2, leaving me to integrate 1/(x^2 -0.25)^0.5 to get 0.5ln(x+(x^2 -0.25)) . However, I keep getting an answer of 0.5(ln(5/2) -ln(1)) when I sub in 1.3 and 0.625. The mark scheme hasn't taken the 1/2 out and has instead integrated to get 0.5ln(2x+(4x^2 -1)^0.5) which I understand. I just don't understand why my method isn't working.
Any help would be appreciated
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ghostwalker
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(Original post by lilybeth647321)
Hi im having a problem with this question:
(iii) Show that the integral of 1/(4x^2 -1)^0.5 (with lower bound 0.625 and upper bound 1.3) = 0.5(ln(5)-ln(2))

I took out a multiple of 1/2, leaving me to integrate 1/(x^2 -0.25)^0.5 to get 0.5ln(x+(x^2 -0.25)) . However, I keep getting an answer of 0.5(ln(5/2) -ln(1)) when I sub in 1.3 and 0.625. The mark scheme hasn't taken the 1/2 out and has instead integrated to get 0.5ln(2x+(4x^2 -1)^0.5) which I understand. I just don't understand why my method isn't working.
Any help would be appreciated
Your answer is the same as theirs, just a different form.

Use the "laws of logs" to simplify your answer. Don't forget ln(1) is 0.
Last edited by ghostwalker; 3 weeks ago
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lilybeth647321
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(Original post by ghostwalker)
Your answer is the same as theirs, just a different form.

Use the "laws of logs" to simplify your answer. Don't forget ln(1) is 0.
of course! Thanks for helping out
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