What is the definite integral int_0^1 dx/(1+sqrt x)^4
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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`
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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