Question

How do you evaluate `int [x / sqrt (2x-1)]dx` for [1, 5]?

Answer 1

`16/3`.

Explanation:

Let, `I=int_1^5x/sqrt(2x-1)dx`.

`:. I=1/2int_1^5{(2x-1)+1}/sqrt(2x-1)dx`,

`=1/2int_1^5{(2x-1)/sqrt(2x-1)+1/sqrt(2x-1)}dx`,

`=1/2int_1^5{sqrt(2x-1)+1/sqrt(2x-1)}dx`.

Now, recall that,

`intsqrttdt=2/3t^(3/2)+c_1, &, int1/sqrttdt=2sqrtt+c_2`.

`:. I=1/2[1/2*2/3(2x-1)^(3/2)+1/2*2sqrt(2x-1)]_1^5`,

`=1/2[1/3(2x-1)^(3/2)+sqrt(2x-1)]_1^5`,

`=1/2[{1/3(2xx5-1)^(3/2)+sqrt(2xx5-1)}]`

`-1/2[{1/3(2xx1-1)^(3/2)+sqrt(2xx1-1)}]`,

`=1/2[{1/3*3^3+3}-{1/3*1+1}]`,

`=1/2[12-4/3]`.

`I=16/3`.

Answer 2

`int_1^5 x/sqrt(2x-1) dx = 16/3`

Explanation:

`I = int_1^5 x/sqrt(2x-1) dx`

Let `u = 2x-1`

`-> x=(u+1)/2 and dx = (du)/2`

Indefinite `I = int ((u+1)/2)/sqrt(u) (du)/2`

`= 1/4 int (u+1)/sqrtu du`

`= 1/4 int(u^(1/2) + u^(-1/2)) du`

Apply linearity

`= 1/4 int u^(1/2)du + int u^(-1/2) du`

Apply power rule

`= 1/4 [u^(3/2)/(3/2) + u^(1/2)/(1/2)] +C`

`= [u^(3/2)/6 +u^(1/2)/2]+C`

Undo substitution

`I= [(2x-1)^(3/2)/6 + (2x-1)^(1/2)/2]_1^5`

`= [(2(2x-1)^(3/2)+6(2x-1)^(1/2))/12]_1^5`

`= [(sqrt(2x-1)(2(2x-1)+6})/12]_1^5`

`= [(sqrt(2x-1)(4x-2+6))/12]_1^5`

`= [(sqrt(2x-1)(x+1})/3]_1^5`

`= ((sqrt(9)xx6)/3 - (sqrt(1)xx2)/3)`

`=(6-2/3) = 16/3`

Answer 3

`16/3`

Explanation:

`int_{1}^{5}x/sqrt(2x-1)dx`

Let `u = 2x-1`

Then, `2x = u+1`

`\thereforex=(u+1)/2`

Using this, we can say that:

`(dx)/(du) = 1/2`

`\thereforedx=1/2du`

The reason why we did all this calculation is so that we can turn

`int_1^5x/sqrt(2x-1)dx`

into a simpler integral of:

`int_1^9(u+1)/(2sqrt(u))*1/2du`

Notice that the ranges have changed. This is because the variable have changed to `u` instead of `x`. So the values have to change to be in terms of `u` not `x`.

`= 1/2int_1^91/2*(u+1)/(sqrt(u))du`

`= 1/4int_1^9(u+1)/(sqrt(u))du`

`= 1/4int_1^9u/(sqrt(u)) + 1/sqrt(u)du`

`= 1/4int_1^9u/(sqrt(u))du + 1/4int_1^91/(sqrt(u))du`

`= 1/4int_1^9u^(1/2)du + 1/4int_1^9u^(-1/2)du`

`= 1/4[2/3u^(3/2)]_1^9 + 1/4[2u^(1/2)]_1^9`

Assuming that you know how to do basic integrals, I will skip the steps to solve the above expression.

After solving the above:

`= 13/3 + 1`

`= 16/3`

Hope that makes sense! If it doesn't make sense, feel free to comment any questions.