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junior college 2 | H1 Maths
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junior college 2 chevron_right H1 Maths chevron_right Singapore

Need help, why is the first method wrong
∫(upper limit: 0, lower limit: -2) 1/2x^2 + 5dx

Date Posted: 2 years ago
Views: 268
J
J
2 years ago
Realise that you have a constant ½ outside, but your 5x originally didn't have it.
J
J
2 years ago
∫(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
J
J
2 years ago
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
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2 years ago
Thank you for explaining!

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