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junior college 2 | H2 Maths
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Could some one help me solve qn f) and 4 without gc:-( asap! I need to submit it really soon! And I'm at wits end!!! Send help please
Date Posted: 4 years ago
Views: 338
Hii, may I ask where/what year this question is from? Cause for Qns 4 I am able to integrate the front and back integrals but the middle integral requires a certain formula or method (that I didn't learn during my JC years) to get 22/7 - π
This is actually his assignment for Introductory Mathematics MA1301, one of the math modules in NUS designed to bring students up to par with A level standard of Math proficiency . He posted the same question some time earlier
The middle integral uses long division (see my working in the comments under your posted solution)
The middle integral uses long division (see my working in the comments under your posted solution)
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done
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For (f)...not sure whether this is correct/the most efficient method since it's been a long time since I've done integration
Date Posted:
4 years ago
Slight error in the simultaneous equation solving
① - ② :
A + 3C - (A - C) = -2 - 2
4C = -4
C = - 1
Then A = C + 2
= -1 + 2
= 1
A + 3C - (A - C) = -2 - 2
4C = -4
C = - 1
Then A = C + 2
= -1 + 2
= 1
Final expression :
ln(x + 1) - 1/(x + 1) - ln(x - 1) - 1/(x - 1)
= ln((x+1)/(x - 1)) - (x + 1 + x - 1)/(x + 1)(x-1))
= ln[(x+1)/(x-1)] - 2x/(x² - 1) + constant
ln(x + 1) - 1/(x + 1) - ln(x - 1) - 1/(x - 1)
= ln((x+1)/(x - 1)) - (x + 1 + x - 1)/(x + 1)(x-1))
= ln[(x+1)/(x-1)] - 2x/(x² - 1) + constant
好心人
Oh rightt I see the error. Thank u J
好有心
Welcome. Good effort anyway.
true
Thank u so much
done
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This is all I can offer for Q4 huhu, my brain can't integrate the middle to get 22/7 - π
Date Posted:
4 years ago
Wow much effort..salute!
Actually can use binomial theorem to expand :
∫ x⁴(1 - x)⁴ dx
= ∫ ( x⁴ (1⁴ + 4C1(1³(-x)) + 4C2(1²(-x)²) + 4C3(1(-x)³ + x⁴) ) dx
= ∫ (x⁴ (1 - 4x + 6x² - 4x³ + x⁴) ) dx
= ∫ ( x⁴ - 4x^5 + 6x^6 - 4x^7 + x^8) dx
= 1/5 x^5 - ⅔x^6 + 6/7 x^7 - ½ x^8 + 1/9 x^9 + c
Actually can use binomial theorem to expand :
∫ x⁴(1 - x)⁴ dx
= ∫ ( x⁴ (1⁴ + 4C1(1³(-x)) + 4C2(1²(-x)²) + 4C3(1(-x)³ + x⁴) ) dx
= ∫ (x⁴ (1 - 4x + 6x² - 4x³ + x⁴) ) dx
= ∫ ( x⁴ - 4x^5 + 6x^6 - 4x^7 + x^8) dx
= 1/5 x^5 - ⅔x^6 + 6/7 x^7 - ½ x^8 + 1/9 x^9 + c
Haha thank u. Usually avoid binomial theorem for expansion cause of the probability of making error on my part~
Middle integral is done using long division :
∫¹ ,0 x⁴(1-x)⁴ / (1 + x²) dx
= ∫¹ ,0 (x⁴ - 4x^5 + 6x^6 - 4x^7 + x^8)/(1 + x²) ) dx
= ∫¹,0 (x^6 - 4x^5 + 5x⁴ - 4x² + 4 - 4/(x² + 1) ) dx
= [ 1/7 x^7 - ⅔x^6 + x^5 - 4/3 x³ + 4x - 4tan-¹ x ]¹,0
= (1/7 - ⅔ + 1 - 4/3 + 4 - 4(π/4) )
= 22/7 - π
= 3 1/7 - π
∫¹ ,0 x⁴(1-x)⁴ / (1 + x²) dx
= ∫¹ ,0 (x⁴ - 4x^5 + 6x^6 - 4x^7 + x^8)/(1 + x²) ) dx
= ∫¹,0 (x^6 - 4x^5 + 5x⁴ - 4x² + 4 - 4/(x² + 1) ) dx
= [ 1/7 x^7 - ⅔x^6 + x^5 - 4/3 x³ + 4x - 4tan-¹ x ]¹,0
= (1/7 - ⅔ + 1 - 4/3 + 4 - 4(π/4) )
= 22/7 - π
= 3 1/7 - π
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