I have the equation for the top half of an ellipse:
$$f(s) = \frac{2}{3\pi}\times \sqrt{1-\bigg(\frac{s-5}{3}\bigg)^2}$$
For one of my classes I have to find the integral from $2$ to $x$ of this function but I haven't learned integrals yet so I have no idea what to do. I think this is called a definite integral but again I am lost because I haven't learned about integrals.
Hopefully you know that if we have a function $f(x)=x^n$ then we can differentiate this to give us $f'(x)=nx^{n-1}$
Integration is the opposite to this, in the same way that subtraction is the opposite to addition, we can say that $$\int f'(x)\,dx= \int nx^{n-1}\,dx=x^n+C$$
We must add the $+C$ as the original $f(x)$ could have had a constant which would have been turned to zero when it was differentiated $\left(f'(x^n+1)=f'(x^n+2)=\cdots =f'(x^n+C)=nx^{n-1}\right)$
We can then perform a definite integral as follows, for $a<b$:
\begin{align}\int_a^b f'(x)\,dx &= \int_a^b nx^{n-1}\,dx\\ &= \left[x^n+C\right]_{x=b}-\left[x^n + C\right]_{x=a}\\ &= b^n+C-a^n-C\\ &= b^n-a^n\end{align}
Does this help you solve your problem?