# Relating integrals to a square

## The following problem is a problem in my linear algebra book. (a) Compute the following integrals (you can use a calculator) and state the relationship between them. $$\int_{-1}^{1} x^2\mathop{dx} \hspace{1em} \int_{-1}^{1} (x^2 - 1/3)^2 \mathop{dx} \hspace{1em} \int_{-1}^{1} (x^2 + x - 1/3)^2 \mathop{dx}$$ I evaluated the integrals and got $2/3, 8/45$, and $38/45$ respectively. The relationship is that the sum of the first two integrals equals the second. (b) Evaluate the integral $$\int_{-1}^{1} x(x^2 - 1/3) \mathop{dx}$$ I evaluated the integral to be $0$. (c) Explain how these results relate to the fact that the diagonal of a unit square is equal to $\sqrt{2}$ units. I'm not sure how to do (c) at all. I tried graphing the functions etc, but I can't find any relationship between these integrals and a square. Can someone please help me here? The following problem is a problem in my linear algebra book.

(a) Compute the following integrals (you can use a calculator) and state the relationship between them.

$$\int_{-1}^{1} x^2\mathop{dx} \hspace{1em} \int_{-1}^{1} (x^2 - 1/3)^2 \mathop{dx} \hspace{1em} \int_{-1}^{1} (x^2 + x - 1/3)^2 \mathop{dx}$$

I evaluated the integrals and got $$2/3, 8/45$$, and $$38/45$$ respectively. The relationship is that the sum of the first two integrals equals the second.

(b) Evaluate the integral

$$\int_{-1}^{1} x(x^2 - 1/3) \mathop{dx}$$

I evaluated the integral to be $$0$$.

(c) Explain how these results relate to the fact that the diagonal of a unit square is equal to $$\sqrt{2}$$ units.

I'm not sure how to do (c) at all. I tried graphing the functions etc, but I can't find any relationship between these integrals and a square. Can someone please help me here?