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

crmhaske | College Teacher | (Level 3) Associate Educator

Posted on

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

`intsqrt(u+1+2sqrt(u))du`

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

`2intvsqrt(v^2+2v+1)dv`

Step 5: Complete the square

`2intvsqrt((v+1)^2)dv=2intv(v+1)dv`

Step 5: Expand

`2int(v^2+v)dv`

Step 6: Separate the terms and integrate

`2intv^2dv+2intvdv`

`2/3v^3+v^2`

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

`2/3u^(3/2)+u`

Step 8: substitute u=x-1 back in

`f(x)=2/3(x-1)^(3/2)+x-1+C`

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

`f(0)=2/3(0-1)^(3/2)+0-1=-2/3i-1`

`f(15)=2/3(15-1)^(3/2)+15-1=48.92`

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

`int^15_0sqrt(x+2sqrt(x-1))=f(15)-f(0)`

`=48.92+2/3i+1=49.92+2/3i`

Sources:

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

We have

`int_0^15sqrt(x+2sqrt(x-1))dx`

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

`so`

`x=t^2+1`

`dx=2tdt`

`x=15,t=sqrt14`

`x=0,t=sqrt(-1)=i`

Thus

`I=int_0^15sqrt(x+2sqrt(x-1))dx=int_i^sqrt(14)sqrt(t^2+1+2t)(2tdt)`

`=2int_i^sqrt(14)tsqrt((t+1)^2)dt`

`=2int_i^sqrt(14)(t^2+t)dt`

`=2{t^3/3+t^2/2}_i^sqrt(14)`

`=2((sqrt(14))^3/3+14/2-i^3/3-i^2/2)`

`=2(17.461+7+i/3+.5)`

`=2(24.962+i/3)`

`Ans.`

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