1)1∫0√x(x − 1)2dx

The integral you're asking about is a bit complex, but it can be solved using the method of substitution. Here's a step-by-step solution:

Step 1: Let's set u = x - 1. Then, du = dx.

Step 2: When x = 0, u = -1. When x = 1, u = 0. So, the limits of the integral change from 0 to 1 to -1 to 0.

Step 3: Substitute u and du into the integral. The integral becomes -∫ from -1 to 0 of √(u+1) * u² du (the negative sign comes from changing the order of the limits of integration).

Step 4: Now, let's set v = u + 1. Then, dv = du.

Step 5: When u = -1, v = 0. When u = 0, v = 1. So, the limits of the integral change from -1 to 0 to 0 to 1.

Step 6: Substitute v and dv into the integral. The integral becomes ∫ from 0 to 1 of √v * (v-1)² dv.

Step 7: This integral can be solved by expanding the (v-1)² term and then integrating term by term.

The integral ∫ from 0 to 1 of √v * (v-1)² dv = ∫ from 0 to 1 of (v^(5/2) - 2v^(3/2) + v^(1/2)) dv.

Step 8: Now, integrate term by term.

= [2/7 * v^(7/2) - 4/5 * v^(5/2) + 2/3 * v^(3/2)] from 0 to 1.

Step 9: Substitute the limits of the integral.

= [2/7 - 4/5 + 2/3] - 0 = -6/35.

So, the value of the original integral ∫ from 0 to 1 of √x * (x - 1)² dx is -6/35.