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Stucked at the evaluate part !!! Helppp
Date Posted: 4 years ago
Views: 411
Quite simple actually. Just realise the following :
8x /( 7√(x-1) )
= 8/7 ( 1/√(x-1) )
= 3/2 × 16/21 × ( 1/√(x-1) )
= 16/21 × ( 3x / (2√(x-1)) )
(How to get 16/21 :
8/7 ÷ 3/2 = 8/7 x 2/3 = 16/21
So 16/21 x 3/2 = 8/7 )
8x /( 7√(x-1) )
= 8/7 ( 1/√(x-1) )
= 3/2 × 16/21 × ( 1/√(x-1) )
= 16/21 × ( 3x / (2√(x-1)) )
(How to get 16/21 :
8/7 ÷ 3/2 = 8/7 x 2/3 = 16/21
So 16/21 x 3/2 = 8/7 )
So,
(upper limit 5, lower limit 2) ∫ 8x/( 7√(x-1) ) dx
= (upper limit 5, lower limit 2) ∫ 16/21 ( 3x / (2√(x-1)) ) dx
= 16/21 (upper limit 5, lower limit 2) ∫ 3x / (2√(x-1)) ) dx
= 16/21 [√(x - 1) (x + 2) ] (upper limit 5, lower limit 2)
Easy t continue from here
(upper limit 5, lower limit 2) ∫ 8x/( 7√(x-1) ) dx
= (upper limit 5, lower limit 2) ∫ 16/21 ( 3x / (2√(x-1)) ) dx
= 16/21 (upper limit 5, lower limit 2) ∫ 3x / (2√(x-1)) ) dx
= 16/21 [√(x - 1) (x + 2) ] (upper limit 5, lower limit 2)
Easy t continue from here
Thankiewwww
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Trick for these questions is to multiply by whatever fraction is required and it’s inverse fraction.
Date Posted:
4 years ago
And it’s reciprocal fraction** my bad.
Tq so much
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