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junior college 1 | H2 Maths
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good afternoon? How do i solve this? (Just part a will do) :)
Date Posted: 3 years ago
Views: 330
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u = e^x
du/dx = e^x = u
1/u du = dx
When x = 0, u = e^0 = 1
When x = ln√3, u = e^(ln√3) = √3
The lower and upper bounds have been changed.
So,∫(lower 0, upper ln√3) e^(3x) / (e^(2x) + 1) dx
= ∫(lower 0, upper ln√3) (e^x)³ / ((e^x)²+ 1) dx
= ∫(lower 1, upper √3) u³/(u²+1) × (1/u) du
= ∫(lower 1, upper √3) u²/(u²+1) du
= ∫(lower 1, upper √3) (u² + 1 - 1)/(u²+1) du
= ∫(lower 1, upper √3) (1 - 1/(u²+1) ) du
= [u - tan-¹u] (lower 1, upper √3)
= [√3 - tan-¹√3] - [1 - tan-¹ 1]
= √3 - π/6 - 1 + π/4
= √3 - 1 - π/12
du/dx = e^x = u
1/u du = dx
When x = 0, u = e^0 = 1
When x = ln√3, u = e^(ln√3) = √3
The lower and upper bounds have been changed.
So,∫(lower 0, upper ln√3) e^(3x) / (e^(2x) + 1) dx
= ∫(lower 0, upper ln√3) (e^x)³ / ((e^x)²+ 1) dx
= ∫(lower 1, upper √3) u³/(u²+1) × (1/u) du
= ∫(lower 1, upper √3) u²/(u²+1) du
= ∫(lower 1, upper √3) (u² + 1 - 1)/(u²+1) du
= ∫(lower 1, upper √3) (1 - 1/(u²+1) ) du
= [u - tan-¹u] (lower 1, upper √3)
= [√3 - tan-¹√3] - [1 - tan-¹ 1]
= √3 - π/6 - 1 + π/4
= √3 - 1 - π/12
Sorry, correction in the 2nd last step.
Should be √3 - π/3 - 1 + π/4
Should be √3 - π/3 - 1 + π/4
Thank you! :)
done
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Hope this helps!
Date Posted:
3 years ago
Thank you for the help! :D
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