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What about:   `int_0^15 sqrt(x+2sqrt(x-1)) dx`  ?

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oldnick | (Level 1) Valedictorian

Posted May 18, 2013 at 4:32 AM via web

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What about:


`int_0^15 sqrt(x+2sqrt(x-1)) dx`  ?

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crmhaske | College Teacher | (Level 3) Associate Educator

Posted May 18, 2013 at 5:36 AM (Answer #1)

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Use integration by substitution

Step 1: define u

`u=x-1` -> `du=dx` and `x=u+1`

Step 2: substitute u and du into the equation


Step 3: define v

`v=sqrt(u)` -> `dv=1/(2sqrt(u))du` and `u=v^2`

`dv=1/(2sqrt(v^2))du=1/(2v)` -> `du=2vdv`

Step 4: Substitute v and dv into the equation


Step 5: Complete the square


Step 5: Expand


Step 6: Separate the terms and integrate



Step 7: substitute` v=sqrt(u)` back in


Step 8: substitute u=x-1 back in


Step 9: evaluate at x=0 and x=15



Final step: compute f(15)-f(0)




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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted May 18, 2013 at 7:03 AM (Answer #3)

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We have


Let `sqrt(x-1)=t,`

















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