How does one solve equations with integrals? I am starting to study linear algebra, in the first pages of the book appeared a problem that I could not tackle. What polynomial $p$ satisfies $\int_{-1}^{1}p(y)\,dy=0$ and $\int_{-1}^{1}yp(y)\,dy=1$. I have some knowledge of differential and integral calculus, but I never learned how to solve equations with integrals. How are they solved and when are you supposed to learn to solve them ?
 A: The integral equations you've mentioned aren't that hard and can be just solved and conclusions can be made just by using brute force : 
Let $p(y)$ be a polynomial, such: 
$$p(y) = a_ny^n + a_{n-1}y^{n-1} + \dots + a_1y + a_0$$
Then the $2$ given integral equations will be : 
$$\int_{-1}^1p(y)dy = \int_{-1}^1(a_ny^n + a_{n-1}y^{n-1} + \dots + a_1y + a_0)dy = 0 $$ 
$$\Rightarrow$$
$$\bigg[\frac{a_n}{n+1}y^{n+1} + \frac{a_{n-1}}{n}y^{n} + \dots + \frac{a_1}{2}y^2 + a_0 y \bigg]_{-1}^1 = 0$$
$$\text{and}$$
$$\int_{-1}^1yp(y)dy = \int_{-1}^1 y(a_ny^n + a_{n-1}y^{n-1} + \dots + a_1y + a_0)dy =1$$
$$\Leftrightarrow$$
$$\int_{-1}^1 (a_ny^{n+1} + a_{n-1}y^{n} + \dots + a_1y^2 + a_0y)dy=1$$
$$\Leftrightarrow$$
$$\bigg[\frac{a_n}{n+2}y^{n+2} + \frac{a_{n-1}}{n+1}y^{n+1} +\dots +\frac{a_1}{3}y^3 + \frac{a_0}{2}y^2 \bigg]_{-1}^1=1$$
Finally, for the polynomial $p(y)$ and it's coefficients and order, you'll get the system : 
$$\begin{cases} \bigg[\frac{a_n}{n+1}y^{n+1} + \frac{a_{n-1}}{n}y^{n} + \dots + \frac{a_1}{2}y^2 + a_0 y \bigg]_{-1}^1 = 0 \\ \bigg[\frac{a_n}{n+2}y^{n+2} + \frac{a_{n-1}}{n+1}y^{n+1} +\dots +\frac{a_1}{3}y^3 + \frac{a_0}{2}y^2 \bigg]_{-1}^1=1  \end{cases}$$
Can you derive a conclusion from now on ?
A: The first equation is true for any odd function, so you could try $\;p(x)=ax+bx^3\;,\;\;a,b\in\Bbb R\;$,  or something of the like.
But if $\;p\;$ is an odd polynomial, then $\;yp(y)\;$ is an even one, so
$$\int_{-1}^1 yp(y)\,dy=2\int_0^1 yp(y)\,dy$$
Observe carefully the above and I bet you'll be able to come up with some ideas...
A: Since the question doesn't ask for all polynomials, we can just find one, so it makes sense to try looking for the smallest. A constant can't work (why not?), so let's try a linear function. $x$ is odd, so something of the form $p(y) = ay$ seems likely, since the constant term must be zero as well. 
Putting this into the second integral gives $$1=\int_{-1}^1 ay^2 \, dy =a\left[\frac{y^3}{3}\right]_{-1}^1=\frac{2a}{3}.$$
So, $p(y) = \frac{3}{2} y$ satisfies the given equation.
