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secondary 4 | A Maths
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secondary 4 chevron_right A Maths chevron_right Singapore

Does anyone know how to do part b? Thank you

Date Posted: 3 years ago
Views: 194
J
J
3 years ago
∫(1,4) [4 - 3f(x)] dx

= ∫(1,4) 4 dx - ∫(1,4) 3f(x) dx

= [4x] (1,4) + ∫(4,1) 3f(x) dx

= 4(1) - 4(4) + 3∫(4,1) f(x) dx

= 4 - 16 + 3(5)

= 4 - 16 + 15

= 3
J
J
3 years ago
Alternative working :

∫(1,4) [4 - 3f(x)] dx

= ∫(4,1) (3f(x) - 4) dx

= ∫(4,1) 3f(x) dx - ∫(4,1) 4 dx

= 3 ∫(4,1) f(x) dx - [4x] (4,1)

= 3(5) - [4(4) - 4(1)]

= 15 - 16 + 4

= 3

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
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The reason why we can do such operations for this integral is that int (1 0 f(x) dx) represents the area under the graph of f(x) from x = 0 to x = 1 and int (4 1 f(x) dx) represents the area under the graph of f(x) from x = 1 to x = 4. Of course, summing up the areas gives us a total area equivalent to the area under the graph of f(x) from x = 0 to x = 4, which is represented as int (4 0 f(x) dx).