How do you integrate #int (x+1)sqrt(2-x)dx#?

Question

How do you  integrate #int (x+1)sqrt(2-x)dx#?

Charles Anctil · Accepted Answer

#(-2(2-x)^(3/2)(x+3))/5+C#

#I=int(x+1)sqrt(2-x)dx#

Let #u=2-x#. This implies that #du=-dx#. Also note that #-u=x-2#, so #-u+3=x+1#. Then:

#I=int(-u+3)sqrtu(-du)=int(u-3)sqrtudu#

Expanding the square root as #u^(1/2)#:

#I=int(u(u^(1/2))-3u^(1/2))du=int(u^(3/2)-3u^(1/2))du#

Now using #intu^ndu=u^(n+1)/(n+1)+C#:

#I=u^(5/2)/(5/2)-3(u^(3/2)/(3/2))=2/5u^(5/2)-2u^(3/2)#

Factoring and making it look nice:

#I=u^(3/2)(2/5u-2)=(u^(3/2)(2u-10))/5=(2u^(3/2)(u-5))/5#

From #u=2-x#:

#I=(2(2-x)^(3/2)((2-x)-5))/5=(-2(2-x)^(3/2)(x+3))/5+C#

Luke Alvarez · Answer

undefined To integrate $ \int (x+1)\sqrt{2-x} \, dx $, you can use the substitution method. Let $ u = 2-x $. Then, $ du = -dx $.

So, $ dx = -du $, and $ x = 2 - u $.

Substituting these into the integral:

$ \int (2 - u + 1)\sqrt{u} \cdot (-du) $

$ = -\int (3 - u)\sqrt{u} \, du $

Expand the expression:

$ = -\int (3\sqrt{u} - u\sqrt{u}) \, du $

Now integrate each term separately:

$ = -\left(3\int \sqrt{u} \, du - \int u\sqrt{u} \, du\right) $

$ = -\left(3 \cdot \frac{2}{3}u^{3/2} - \frac{2}{5}u^{5/2}\right) + C $

$ = -2u^{3/2} + \frac{2}{5}u^{5/2} + C $

Now, revert $ u $ back to $ x $:

$ = -2(2-x)^{3/2} + \frac{2}{5}(2-x)^{5/2} + C $

This is the integral of $ (x+1)\sqrt{2-x} $.