How do you integrate #int (x-1)sqrt(2-x)dx# from [1,2]?

Question

How do you  integrate #int (x-1)sqrt(2-x)dx# from [1,2]?

Emma Aldridge · Accepted Answer

-4/15

u substitution: u = 2-x du = -dx x - 1 = 1 - u, #int_1^2(x-1)sqrt(2-x)dx = -int_1^0(1-u)sqrt(u)du = -int_1^0sqrt(u)-usqrt(u)du = -int_1^0u^(1/2)-u^(3/2)du = int_0^1u^(1/2)-u^(3/2)du# You have to change the bounds from [1,2] to [1,0] when you u-substitute. You find the bounds for u by plugging x=1 and x=2 into the u=2-x to find u=1 and u=0 (keep the order of u=1 before u=0 because u is negatively correlated to x). From there: #= F(1) - F(0)# where #F(x) = 2/3u^(3/2) - 2/5u^(5/2)# #= [2/3(1)^(3/2)-2/5(1)^(5/2)]-[2/3(0)^(3/2)-2/5(0)^(5/2)]# #= [2/3-2/5]-[0-0]# #= 4/15# You don't need to u-substitute back and you'll get the same answer if you do.

Ezekiel Alexander · Answer

undefined To integrate $ \int_{1}^{2} (x-1) \sqrt{2-x} \, dx $ from $[1,2]$, we can use the substitution method. Let's set $ u = 2-x $, then $ du = -dx $. Also, when $ x = 1 $, $ u = 2 - 1 = 1 $, and when $ x = 2 $, $ u = 2 - 2 = 0 $.

Substitute these into the integral:

$$ \int_{1}^{2} (x-1) \sqrt{2-x} \, dx = -\int_{1}^{2} u \sqrt{u} \, du $$

Now we integrate $ -u \sqrt{u} $ with respect to $ u $:

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

Substitute back $ u = 2-x $ and the limits:

$$ -\frac{2}{5}(2-x)^{5/2} \Bigg|_{1}^{2} $$

Now plug in the upper limit:

$$ -\frac{2}{5}(2-2)^{5/2} = 0 $$

And subtract the result when plugging in the lower limit:

$$ 0 - \left(-\frac{2}{5}(2-1)^{5/2}\right) = -\frac{2}{5} $$

So, $ \int_{1}^{2} (x-1) \sqrt{2-x} \, dx = -\frac{2}{5} $.