Find three polynomials $f_1(t),f_2(t),f_3(t)$ of degree at most 2, with real coefficients, such that... $$\int_{0}^{1}f_i(t)f_j(t)dt= \bigg\{ \begin{array}{cc}1,&i=j;\\0,&i\neq j.\end{array}$$
I'm looking for an algorithmic approach to this type of problem. Just beginning the process of brushing up on my calculus, so I don't recall many facts/techniques that will probably help. For example, what it actually "means" for a definite integral to evaluate to $0$. Thus far, my attempt at this problem has been guess-and-check, which is obviously not the best method. 
For example, $f_1(t)=\sqrt{3}t$ satisfies the required properties. So now we want to construct $f_2(t)$. Let's try to find a degree one polynomial that satisfies the required properties. That is, let $f_2(t)=at$, for some $a\in\mathbb{R}$. Then we need $\int_{0}^{1}f_1(t)f_2(t)=0$. This implies that $\frac{a\sqrt{3}}{3}=0$, implying that $a=0$, and hence $f_2(t)=0$. This clearly does not work, as the integral of $0^2$ is $0$, not $1$. So now we look for a degree two polynomial to satisfy the desired properties...etc.
 A: Thanks to the comments and a Matrix Theory textbook by Sylvia A. Hobart, I have a solution.
Define $\mathcal{C}[a,b]$ to be the set of real valued functions that are continuous on the closed interval $[a,b]$. For this to be a vector space, we need addition and scalar multiplication to be defined. Thus, let $f$ and $g$ be functions and define $[f+g](t)=f(t)+g(t)$ and $[rf](t)=r(f(t))$.
Now we define the inner product, $\sigma$, on $\mathcal{C}[a,b]$ by $\sigma(f,g)=\int_{0}^{1}f(t)g(t)dt$. In interest of saving space, we omit verification that $\sigma$ is actually an inner product, but an explanation of this can be found in Hobart, pg. 262-263 (2016 edition).
We are looking for three functions that are orthonormal in $\mathcal{C}[a,b]$ (recall that orthonormal means they are normal; length $1$ with respect to some norm, and orthogonal; their inner product is $0$). Thus, define $g_1(t)=1,g_2(t)=t,g_3(t)=t^2\in\mathcal{C}[a,b]$. We'll use the Gram-Schmidt algorithm to find an orthonormal set of functions, $f_1,f_2,f_3$ that span the same subspace as $g_1,g_2,g_3$. The general Gram-Schmidt process is as follows:
Suppose $v_1,v_2,\cdots,v_k$ are linearly independent vectors in $(V,\beta)$, an inner product space.


*

*Let $q_i=\frac{1}{||v_1||_\beta}v_1$.

*For $i=2$ to $k$, repeat steps 3 and 4.

*Let $w_i=v_i-\beta(q_1,v_i)q_1-\beta(q_2,v_i)q_2-\cdots-\beta(q_{i-1},v_i)q_{i-1}$.

*Let $q_i=\frac{1}{||w_i||_\beta}w_i$.


In our case, $\beta$ is the inner product, $\sigma$, and the $\beta$ norm is $\sqrt{\sigma}$. Also, $v_i=g_i(t)$, for $i=1,2,3$. Thus, 
$$q_1(t)=\frac{1}{\sqrt{\sigma(g_1(t),g_1(t))}}g_1(t)=\frac{1}{\sqrt{\int_{0}^{1}1^2dt}}=1$$
$$w_2(t)=g_2(t)-\sigma(q_1(t),g_2(t))q_1(t)=t-\int_{0}^{1}tdt=t-\frac{1}{2}$$
$$q_2(t)=\frac{1}{\sqrt{\sigma(w_2(t),w_2(t))}}w_2(t)=\frac{1}{\sqrt{\int_{0}^{1}(t-\frac{1}{2})^2}dt}(t-\frac{1}{2})=\sqrt{3}(2t-1)$$
$$w_3(t)=g_3(t)-\sigma(q_1(t),g_3(t))q_1(t)-\sigma(q_2(t),g_3(t))q_2(t)=t^2-\int_{0}^{1}t^2dt-(\int_{0}^{1}\sqrt{3}(2t-1)t^2dt)(\sqrt{3}(2t-1))=t^2-t+\frac{1}{6}$$
$$q_3(t)=\frac{1}{\sqrt{\sigma(w_3(t),w_3(t))}}w_3(t)=\frac{1}{\sqrt{\int_{0}^{1}(t^2-t+\frac{1}{6})^2dt}}(t^2-t+\frac{1}{6})=\sqrt{5}(6t^2-6t+1)$$
Thus, $f_1(t)=q_1(t), f_2(t)=q_2(t), f_3(t)=q_3(t)$ is our solution.
