integrate (sqrt(2x) - sqrt(2x - x ^ 2)) dx from 0 to 2 = integrate (1 - sqrt(1 - y ^ 2) - (y ^ 2)/2) dy from 0 to 1 + integrate (2 - (y ^ 2)/2) dy from 1 to 2
Hello! I hope you're doing well. If you have any questions about colleges or courses, feel free to ask, and I'd be happy to help you out!
As for the integral you mentioned, I'd suggest breaking it down step by step:
- The first part:
∫(2x−2x−x2) dxfrom 0 to 2\int (\sqrt{2x} - \sqrt{2x - x^2}) \, dx \quad \text{from} \, 0 \, \text{to} \, 2∫(2x−2x−x2)dxfrom0to2
- The second part:
∫(1−1−y2−y22) dyfrom 0 to 1\int \left(1 - \sqrt{1 - y^2} - \frac{y^2}{2}\right) \, dy \quad \text{from} \, 0 \, \text{to} \, 1∫(1−1−y2−2y2)dyfrom0to1
and then
∫(2−y22) dyfrom 1 to 2.\int \left(2 - \frac{y^2}{2}\right) \, dy \quad \text{from} \, 1 \, \text{to} \, 2.∫(2−2y2)dyfrom1to2.
If you'd like help solving this or discussing it further, let me know! Otherwise, feel free to ask about anything else you'd like assistance with.
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