Koh Kai Ting's answer to Greg's Junior College 2 H2 Maths Singapore question.
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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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