Summary

The core idea. For $f(x)\geq 0$ on $[a,b]$, the definite integral $\int_a^b f(x)\,dx$ equals the area bounded by $y=f(x)$, the $x$-axis, and the ordinates $x=a,\,x=b$ -- justified by the Riemann-sum limit of thin rectangular strips. For the region between two curves with $f(x)\geq g(x)$ on $[p,q]$,

Area under a line $y=mx+c$ from $x=a$ to $x=b$ (line staying above the $x$-axis):

matching the elementary trapezium formula, where $y_1,y_2$ are the line's heights at $x=a,x=b$.

Area of a circle $x^2+y^2=r^2$: quarter-circle area $\dfrac{\pi r^2}{4}$ (via $x=r\sin\theta$), so

Useful antiderivative: $\displaystyle\int\sqrt{r^2-x^2}\,dx=\frac{x}{2}\sqrt{r^2-x^2}+\frac{r^2}{2}\sin^{-1}\!\left(\frac{x}{r}\right)+C$.

Area under a parabola $y^2=4ax$ and the line $x=h$ (region symmetric about the $x$-axis):

For a region bounded by the $y$-axis and $y=k$: $\text{Area}=\displaystyle\int_0^k \frac{y^2}{4a}\,dy=\frac{k^3}{12a}$.

Area of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$: an ellipse is a circle of radius $a$ scaled vertically by $b/a$, so its quarter-area is $\dfrac{b}{a}\cdot\dfrac{\pi a^2}{4}=\dfrac{\pi ab}{4}$, giving

Setting $a=b=r$ recovers the circle formula, a built-in consistency check. …