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junior college 2 | H1 Maths
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Need help, why is the first method wrong
∫(upper limit: 0, lower limit: -2) 1/2x^2 + 5dx
Date Posted: 4 years ago
Views: 441
Realise that you have a constant ½ outside, but your 5x originally didn't have it.
∫(lower -2, upper 0) (½x² + 5) dx
= ∫(lower -2, upper 0) ½x² dx + ∫(lower -2, upper 0) 5 dx
= [(⅓)½x³] - [5x]
= [1/6 (0³) - 1/6 (-2)³] - [5(0) - 5(2)]
= [-1/6 (-8) ] - [-10]
= 4/3 + 10
= 11⅓
= ∫(lower -2, upper 0) ½x² dx + ∫(lower -2, upper 0) 5 dx
= [(⅓)½x³] - [5x]
= [1/6 (0³) - 1/6 (-2)³] - [5(0) - 5(2)]
= [-1/6 (-8) ] - [-10]
= 4/3 + 10
= 11⅓
What you were probably trying to do was this :
∫(lower -2, upper 0) (½x² + 5) dx
= ½∫(lower -2, upper 0) x² dx + ∫(lower -2, upper 0) 5 dx
= ½[⅓x³] + [5x]
= ½[⅓x³] + ½[10x]
= ½ [⅓x³ + 10x]
= ½ [ (⅓(0³) + 10(0)) - (⅓(-2)³ + 10(-2)) ]
= ½[-(-8/3 - 20)]
= ½ (8/3 + 20)
= ½ (22⅔)
= 11⅓
∫(lower -2, upper 0) (½x² + 5) dx
= ½∫(lower -2, upper 0) x² dx + ∫(lower -2, upper 0) 5 dx
= ½[⅓x³] + [5x]
= ½[⅓x³] + ½[10x]
= ½ [⅓x³ + 10x]
= ½ [ (⅓(0³) + 10(0)) - (⅓(-2)³ + 10(-2)) ]
= ½[-(-8/3 - 20)]
= ½ (8/3 + 20)
= ½ (22⅔)
= 11⅓
Thank you for explaining!
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