Evaluate the integral: I = integral sqrt(2+x-x²) dx
Evaluate the integral: $$ I = \int \sqrt{2+x-x^2}\,dx. $$
Explanation of Method
Completing the square converts the radicand into the bounded $a^2-u^2$ form, which must be resolved with the arcsine formula $\int\sqrt{a^2-u^2}\,du=\tfrac u2\sqrt{a^2-u^2}+\tfrac{a^2}2\arcsin(u/a)$ — not the logarithmic formula that applies to $u^2\pm a^2$.
- Complete the square for $2+x-x^2$.
- Substitute $u=x-\tfrac12$.
- Apply the standard formula for $\int\sqrt{a^2-u^2}\,du$.
- Substitute back.
Step 1 · Complete the square
$$ 2+x-x^2 = -(x^2-x-2) = \frac94 - \left(x-\frac12\right)^2. $$
Step 2 · Substitute
Let $u=x-\tfrac12$, $du=dx$, $a=\tfrac32$:
$$ I = \int \sqrt{a^2-u^2}\,du. $$
Step 3 · Apply the standard arcsine-based formula
$$ I = \frac u2\sqrt{a^2-u^2} + \frac{a^2}2\arcsin\frac ua + C. $$
Step 4 · Substitute back
With $u=x-\tfrac12$, $a=\tfrac32$, so $\tfrac u2=\tfrac{2x-1}4$ and $\tfrac ua=\tfrac{2x-1}3$:
$$ I = \frac{2x-1}{4}\sqrt{2+x-x^2} + \frac98\arcsin\left(\frac{2x-1}{3}\right) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Completed square | $\tfrac94-(x-\tfrac12)^2$ |
| $u,\,a$ | $x-\tfrac12,\ \tfrac32$ |
Additional Notes
- Correction: the source applied the logarithmic identity for $u^2-a^2$-type radicals to a genuinely bounded $a^2-u^2$-type radical. Since $2+x-x^2$ is a downward-opening parabola (real, finite domain $-1\leq x\leq2$), the correct antiderivative formula uses $\arcsin$, not $\ln$; the source's stated answer was never actually evaluated to a concrete closed form (it referenced "the identity" without applying it correctly).
- The domain requires $2+x-x^2\geq0$, i.e. $-1\leq x\leq2$ (the roots of $-(x-2)(x+1)$).
- The result was verified by differentiation: writing the product-rule and arcsine-rule derivatives of $\tfrac{2x-1}4\sqrt{2+x-x^2}+\tfrac98\arcsin\!\left(\tfrac{2x-1}3\right)$ over the common denominator $8\sqrt{2+x-x^2}$ (using $9-(2x-1)^2=4(2+x-x^2)$) gives numerator $4(2+x-x^2)+4(2+x-x^2)=8(2+x-x^2)$, which reduces exactly to $\sqrt{2+x-x^2}$.
Result
The solution to the integral is: