Quick questions on Area between two curves explained: O-Level A-Maths

The vertical gap between two graphs at each $x$ is (upper function) minus (lower function). Adding up these gaps across the region gives the enclosed area:

The limits $a$ and $b$ are the $x$-coordinates where the two graphs intersect. Set the two expressions equal, solve the resulting equation, and use the solutions as the limits.

Because you subtract before integrating, the whole region is handled by one definite integral; you do not need to compute two separate areas and subtract them, although that also works.

Some regions are far easier to integrate horizontally, treating $x$ as a function of $y$. When the boundaries are naturally written as $x = \mathrm{f}(y)$ (for example a sideways parabola), the area becomes $\int_c^d (x_{\text{right}} - x_{\text{left}})\,dy$, the mirror image of the usual formula. The limits are then $y$-values of the intersections. Recognising when a region is bounded more simply left-and-right than top-and-bottom, and switching the variable of integration accordingly, can turn an awkward two-part vertical integral into a single clean horizontal one.

The limits are the intersection points; find them by setting the two expressions equal, not by guessing.

State the integral for the area between $y = x^2$ and $y = 4$ from their intersections. [2 marks]

Find where $y = x^2$ meets $y = 2x$. [2 marks]

Find the area between $y = x^2$ and $y = 2x$. [4 marks]