evaluating an integral question?

Question

hi. i would like to find out how to solve this problem.

i would like to evaluate the integral
∫ x + square root 5- x dx

the square root sign is under the 5-x

thanks for any help. it is very much appreciated

Answer 1

First separate the integral into ∫x dx + ∫sqrt(5 - x) dx. ∫x dx is trivial. The part you're having trouble with is ∫sqrt(5 - x) dx. So, let u = 5 - x. Then du/dx = -1, so dx = -du. Now you have ∫sqrt(5 - x) dx = -∫sqrt(u) du. Note that there is a negative sign in front of the integral. Rewrite this as -∫u^0.5 du = (-u^1.5)/1.5 = (-2/3)u^1.5 = (-2/3)sqrt(u^3), depending on how you prefer to write it. But u = 5 - x and you still need to add the result of the first part of the integral that you separated out. Your final result is (1/2)x^2 - (2/3)sqrt((5 - x)^3) + C.

Answer 2

You solve it by method of substitution.

Let u = 5 - x => x = 5 - u. Then dx = -du.
Substituting the variables, you get
∫ [(5 - u) + sqrt(u)](-du)
=> ∫ [u - 5 - sqrt(u)]du

Integrating this expression, you get
=> (1/2)u^2 - 5u - (2/3)u^(3/2) + C

Substituting back the value of u, the expression becomes
(1/2)(5 - x)^2 - 5(5 - x) - (2/3)(5 - x)^(3/2) + C

You can then simply this algebraically as you please to get the answer.

Answer 3

let U^2= 5-x or x = 5- U^2
2UdU = -dx
dx = -2U dU

∫ x + sqrt(5-x) dx
= ∫ [(5- U^2) + U ]*-2U dU
= -2∫ 5U - U^3 +U^2 dU
=-2[2.25U^2 -0.25U^4 + U^3 /3 ] +C