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junior college 2 | H2 Maths
2 Answers Below
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Please help! Thank you:)
Date Posted: 4 years ago
Views: 742
Integrate by parts,
∫ x√(1+x) dx
= x(1+x)³/² / (3/2) - ∫(1+x)³/² / (3/2) dx
= ⅔x(1+x)³/² - ⅔(1+x)^(5/2) / (5/2)
= ⅔x(1+x)³/² - 4/15 (1+x)^(5/2)
= (2/15)(1+x)³/² (5x - 2(1+x))
= 2/15 (1+x)³/² (3x - 2)
= 2/15 [(1+k)³/² (3k - 2)] - 2/15 [(1+0)³/² (3(0) - 2)]
= 2/15 (1+k)³/² (3k - 2) + 4/15
So 2/15 (1+k)³/² (3k - 2) + 4/15 = 4/15
2/15 (1+k)³/² (4/15 - 2/5 k) = 0
(1+k)³/² = 0 or 4/15 - 2/5 k = 0
(Reject as k is positive constant, so (1+k)³/² > 0 for all real k)
4/15 = 2/5 k
k = 4/15 x 5/2
k = ⅔
∫ x√(1+x) dx
= x(1+x)³/² / (3/2) - ∫(1+x)³/² / (3/2) dx
= ⅔x(1+x)³/² - ⅔(1+x)^(5/2) / (5/2)
= ⅔x(1+x)³/² - 4/15 (1+x)^(5/2)
= (2/15)(1+x)³/² (5x - 2(1+x))
= 2/15 (1+x)³/² (3x - 2)
= 2/15 [(1+k)³/² (3k - 2)] - 2/15 [(1+0)³/² (3(0) - 2)]
= 2/15 (1+k)³/² (3k - 2) + 4/15
So 2/15 (1+k)³/² (3k - 2) + 4/15 = 4/15
2/15 (1+k)³/² (4/15 - 2/5 k) = 0
(1+k)³/² = 0 or 4/15 - 2/5 k = 0
(Reject as k is positive constant, so (1+k)³/² > 0 for all real k)
4/15 = 2/5 k
k = 4/15 x 5/2
k = ⅔
See 2 Answers
Thank u:)
done
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clear
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I would recommend you use integration by parts to approach this question.
Hope this helps!
Hope this helps!
Date Posted:
4 years ago
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