The integral from 1 to 2 of x*the square root of (x-1) dx.
Using the substitution method, I said that u = x-1, du = 1 dx, and dx = du. I'm not sure how to plug u back into the equation since I still have that first x. Suggestions are greatly appreciated.
Let u^2=x-1
2udu=dx
INT (u^2+1)u 2udu=
INT (2u^4+2U^2) du
revise your limits.
To solve this integral using the substitution method, you correctly set u = x - 1, which means du = dx. Now, let's substitute these expressions into the integral:
∫(1 to 2) x*√(x-1) dx
When x = 1, we can solve for u:
u = x - 1
u = 1 - 1
u = 0
When x = 2, we can solve for u:
u = x - 1
u = 2 - 1
u = 1
So, the limits of integration can be rewritten in terms of u:
u = 0 when x = 1
u = 1 when x = 2
Next, we need to express x in terms of u using the substitution u = x - 1:
u = x - 1
u + 1 = x
Now we can rewrite the integral in terms of u:
∫(0 to 1) (u + 1) * √u du
Expanding the expression inside the integral:
∫(0 to 1) u*√u + √u du
Now we can integrate each term separately:
∫(0 to 1) u^(3/2) du + ∫(0 to 1) √u du
Using the power rule, we can integrate the first term:
(2/5)*u^(5/2) evaluated from 0 to 1
Simplifying further:
(2/5)*(1)^(5/2) - (2/5)*(0)^(5/2) + ∫(0 to 1) √u du
(2/5) - 0 + ∫(0 to 1) √u du
Now let's integrate the second term:
(2/5) + (2/3)*u^(3/2) evaluated from 0 to 1
Simplifying further:
(2/5) + (2/3)*(1)^(3/2) - (2/3)*(0)^(3/2)
(2/5) + (2/3) - 0
Finally, combining both terms:
(2/5) + (2/3)
= (6 + 10)/15
= 16/15
Therefore, the value of the integral from 1 to 2 of x*√(x-1) dx is 16/15.