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

3 Answers
Jun 18, 2018

# 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#.

Jun 18, 2018

#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#

Jun 18, 2018

#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.