What is the definite integral int_0^1 dx/(1+sqrt x)^4

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The definite integral `int_0^1 dx/(1+sqrt x)^4` has to be determined.

This can be done using substitution. Start with the indefinite integral `int dx/(1+sqrt x)^4`

Let `1 + sqrt x = y`

`dy/dx = (1/2)*1/(sqrt x)`

= `(1/2)*1/(y-1)`

`dx = 2*(y - 1) dy`

Substituting in the integral `int dx/(1+sqrt x)^4` gives

`int 2*(y - 1)/y^4 dy`

= `2*(int 1/y^3 - 1/y^4 dy)`

= `2*(y^-2/(-2) + y^-3/3)`

As `y = 1 + sqrt x` , this gives:

`-1/(1 + sqrt x)^2 + 2/(3*(1+sqrt x)^3)`

Now taking into accounts the limits we have:

`-(1/(1 + sqrt 1)^2 - 1/(1+sqrt 0)) + 2/(3*(1+sqrt 1)^3) - 2/(3*(1+sqrt 0)^3) `

= `-(1/4 - 1) + 2/24 - 2/3`

= `1/6`

The value of the definite integral `int_0^1 dx/(1+sqrt x)^4 = 1/6`

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kalau | (Level 2) Adjunct Educator

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What is the definite integral int_0^1 dx/(1+sqrt x)^4?

`int_0^1 1/ (1+sqrtx)^4dx`

Let: `U=1+sqrtx`

Then:  `dU= 1/2 x^(-1/2) dx= 1/(2sqrtx) dx`

We can isolate dx by multiplying the 2 square root x on both sides.

`2sqrtx dU = dx`

We also can manipulate the U such that:  `U-1=sqrtx`

And then we have:  `2(U-1) = dx`

Substitute everything back into the integral in terms of U and du.

`int 1/U^4 * 2(U-1) du= 2 int (U-1)/U^4 dU`

Separate the integral.

`2 int U^-3 dU-2int U^-4 dU`

Integrate and simplify:

`2[U^-2 /-2] - 2[U^-3/-3] = -1/U^2 + 2/(3U^3) `

Replace the U term and evaluate each bounds from 0 to 1.

`-[1/ (1+sqrtx)^2 ]_0^1+ [2/(3(1+sqrtx)^3)]_0^1 ` 

`-[1/ (1+sqrt1)^2 - 1/((1+sqrt0)^2)] + [2/ (3(1+sqrt1)^3) - 2/(3(1+sqrt0)^3)]` 

`-[1/4 - 1]+[2/ (3*8) -2/3] = 3/4 + 1/12 -2/3= 9/12+1/12-8/12 =2/12 =1/6`

The answer to this definite integral is:  `1/6`

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