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Question

Question: √(x-x^2) dx...

√(x-x^2) dx

Answer

2x−14x−x2+18sin⁡−1(2x−1)+C\frac{2x - 1}{4}\sqrt{x - x^2} + \frac{1}{8}\sin^{-1}(2x - 1) + C

Explanation

Solution

To evaluate the integral ∫x−x2dx\int \sqrt{x-x^2} dx, we first complete the square for the quadratic expression inside the square root.

  1. Complete the square: The expression is x−x2x - x^2. We can rewrite this as −(x2−x)-(x^2 - x). To complete the square for x2−xx^2 - x, we add and subtract (−12)2=14(\frac{-1}{2})^2 = \frac{1}{4}. So, x2−x=(x2−x+14)−14=(x−12)2−14x^2 - x = (x^2 - x + \frac{1}{4}) - \frac{1}{4} = (x - \frac{1}{2})^2 - \frac{1}{4}. Therefore, x−x2=−[(x−12)2−14]=14−(x−12)2x - x^2 = -[(x - \frac{1}{2})^2 - \frac{1}{4}] = \frac{1}{4} - (x - \frac{1}{2})^2.

  2. Substitute to a standard form: Let u=x−12u = x - \frac{1}{2}. Then du=dxdu = dx. The integral becomes: ∫14−u2du\int \sqrt{\frac{1}{4} - u^2} du This is in the standard form ∫a2−u2du\int \sqrt{a^2 - u^2} du, where a2=14a^2 = \frac{1}{4}, so a=12a = \frac{1}{2}.

  3. Apply the standard integration formula: The formula for ∫a2−x2dx\int \sqrt{a^2 - x^2} dx is x2a2−x2+a22sin⁡−1(xa)+C\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C. Applying this with uu instead of xx and a=12a = \frac{1}{2}: ∫14−u2du=u214−u2+(12)22sin⁡−1(u12)+C\int \sqrt{\frac{1}{4} - u^2} du = \frac{u}{2}\sqrt{\frac{1}{4} - u^2} + \frac{\left(\frac{1}{2}\right)^2}{2}\sin^{-1}\left(\frac{u}{\frac{1}{2}}\right) + C =u214−u2+142sin⁡−1(2u)+C= \frac{u}{2}\sqrt{\frac{1}{4} - u^2} + \frac{\frac{1}{4}}{2}\sin^{-1}(2u) + C =u214−u2+18sin⁡−1(2u)+C= \frac{u}{2}\sqrt{\frac{1}{4} - u^2} + \frac{1}{8}\sin^{-1}(2u) + C

  4. Substitute back to the original variable: Now substitute u=x−12u = x - \frac{1}{2} back into the expression: x−12214−(x−12)2+18sin⁡−1(2(x−12))+C\frac{x - \frac{1}{2}}{2}\sqrt{\frac{1}{4} - \left(x - \frac{1}{2}\right)^2} + \frac{1}{8}\sin^{-1}\left(2\left(x - \frac{1}{2}\right)\right) + C

  5. Simplify the expression:

    • x−122=2x−14\frac{x - \frac{1}{2}}{2} = \frac{2x - 1}{4}
    • 14−(x−12)2=14−(x2−x+14)=14−x2+x−14=x−x2\sqrt{\frac{1}{4} - \left(x - \frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} - (x^2 - x + \frac{1}{4})} = \sqrt{\frac{1}{4} - x^2 + x - \frac{1}{4}} = \sqrt{x - x^2}
    • 2(x−12)=2x−12\left(x - \frac{1}{2}\right) = 2x - 1

    Substituting these simplified terms back, we get: 2x−14x−x2+18sin⁡−1(2x−1)+C\frac{2x - 1}{4}\sqrt{x - x^2} + \frac{1}{8}\sin^{-1}(2x - 1) + C

The integral ∫x−x2dx\int \sqrt{x-x^2} dx is solved by first completing the square of the quadratic expression x−x2x-x^2 to transform it into the form 14−(x−12)2\frac{1}{4} - (x-\frac{1}{2})^2. A substitution u=x−12u = x-\frac{1}{2} converts the integral into the standard form ∫a2−u2du\int \sqrt{a^2-u^2}du, where a=12a=\frac{1}{2}. Using the known integration formula ∫a2−x2dx=x2a2−x2+a22sin⁡−1(xa)+C\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a}) + C, and substituting back u=x−12u=x-\frac{1}{2}, leads to the final result.