1. **Problem statement:** We are given the vector space $P_2$ of polynomials of degree at most 2 with inner product $\langle p,q \rangle = \int_0^1 p(x)q(x)\,dx$. The basis is $S = \{1, x, x^2\}$. We want to apply the Gram-Schmidt process to find an orthonormal basis. 2. **Gram-Schmidt process formula:** Given vectors $v_1, v_2, v_3$, the orthogonal vectors $u_1, u_2, u_3$ are computed as: $$ \begin{aligned} u_1 &= v_1 \\ u_2 &= v_2 - \frac{\langle v_2, u_1 \rangle}{\langle u_1, u_1 \rangle} u_1 \\ u_3 &= v_3 - \frac{\langle v_3, u_1 \rangle}{\langle u_1, u_1 \rangle} u_1 - \frac{\langle v_3, u_2 \rangle}{\langle u_2, u_2 \rangle} u_2 \end{aligned} $$ Then normalize each $u_i$ to get orthonormal vectors $e_i = \frac{u_i}{\|u_i\|}$. 3. **Step 1:** Set $v_1 = 1$, $v_2 = x$, $v_3 = x^2$. Calculate $u_1 = v_1 = 1$. Calculate $\langle u_1, u_1 \rangle = \int_0^1 1 \cdot 1 \, dx = 1$. 4. **Step 2:** Calculate $$ \langle v_2, u_1 \rangle = \int_0^1 x \cdot 1 \, dx = \frac{1}{2} $$ Then $$ u_2 = v_2 - \frac{1/2}{1} \cdot 1 = x - \frac{1}{2}$$ Calculate $\langle u_2, u_2 \rangle = \int_0^1 (x - \frac{1}{2})^2 dx = \int_0^1 (x^2 - x + \frac{1}{4}) dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} + \frac{x}{4} \right]_0^1 = \frac{1}{3} - \frac{1}{2} + \frac{1}{4} = \frac{1}{12}$. 5. **Step 3:** Calculate $$ \langle v_3, u_1 \rangle = \int_0^1 x^2 \cdot 1 \, dx = \frac{1}{3} $$ $$ \langle v_3, u_2 \rangle = \int_0^1 x^2 (x - \frac{1}{2}) dx = \int_0^1 (x^3 - \frac{x^2}{2}) dx = \left[ \frac{x^4}{4} - \frac{x^3}{6} \right]_0^1 = \frac{1}{4} - \frac{1}{6} = \frac{1}{12} $$ Then $$ \nu_3 = v_3 - \frac{1/3}{1} \cdot 1 - \frac{1/12}{1/12} (x - \frac{1}{2}) = x^2 - \frac{1}{3} - 1 \cdot (x - \frac{1}{2}) = x^2 - x + \frac{1}{6} $$ Calculate $\langle u_3, u_3 \rangle = \int_0^1 (x^2 - x + \frac{1}{6})^2 dx = \frac{1}{180}$. 6. **Step 4:** Normalize each vector: $$ e_1 = \frac{u_1}{\sqrt{1}} = 1 $$ $$ e_2 = \frac{u_2}{\sqrt{1/12}} = \sqrt{12} \left(x - \frac{1}{2}\right) = 2\sqrt{3} \left(x - \frac{1}{2}\right) $$ $$ e_3 = \frac{u_3}{\sqrt{1/180}} = \sqrt{180} \left(x^2 - x + \frac{1}{6}\right) = 6\sqrt{5} \left(x^2 - x + \frac{1}{6}\right) $$ **Final orthonormal basis:** $$ \left\{ 1, \quad 2\sqrt{3} \left(x - \frac{1}{2}\right), \quad 6\sqrt{5} \left(x^2 - x + \frac{1}{6}\right) \right\} $$