Evaluate the definite integral. from 4 to 5 of x[root(x-4)] dx?

use u substitution with u = sqrt(x-4)

Then du = 1/2 * (x-4)^(-1/2) * 1 dx = 1/ (2sqrt(x-4))dx

Thus, dx = 2 sqrt(x-4) du and we have:

integral [2 sqrt(x-4) * x / sqrt(x-4)] du. You should notice the sqrt(x-4) cancel out, and we're left with:

integral [2x du]. What do we do now? Recall we let u = sqrt(x - 4) and so u^2 = x - 4, and x = u^2 + 4

So we have: integral [2 (u^2 + 4)du] = 2u^2 + 8 du

After applying integral power rule, we get:

2u^3/3 + 8u evaluated at 4 and 5. Of course we need to remember that u = sqrt(x-4)

So we have:

2/3(root(x-4)) + 8(root(x-4)) evaluated at 5 and 4.

We get:

2/3 root(1) + 8root(1) - (2/3 root0 + 8 root(0))
= 2/3 + 8 - (0 + 0) = 8 + 2/3 or 26/3

Let u = √(x - 4)
Then, u² = x - 4
x = u² + 4 and dx = 2u du

∫x √(x - 4) dx
= ∫(u² + 4) u (2u du)
= ∫ (2u⁴ + 8u²) du
= (2/5)u⁵ + (8/3)u³ + C
= F(x) + C
where F(x) = (2/5)[√(x - 4)]⁵ + (8/3)[√(x - 4)]³

∫_(4 to 5) x √(x - 4) dx
= F(5) - F(4)
= {(2/5)[√(5 - 4)]⁵ + (8/3)[√(5 - 4)]³} - {(2/5)[√(4 - 4)]⁵ + (8/3)[√(4x - 4)]³}
= {(2/5) + (8/3)} - {0 + 0}
= 46/15