Area Under a Curve

The definite integral measures a signed area between a curve and a coordinate axis. For a curve $y=f(x)$ that lies above the $x$-axis on $[a,b]$, the area trapped between the curve, the $x$-axis and the ordinates $x=a$ and $x=b$ is

Think of the region as a stack of vertical strips of width $dx$ and height $y$; each strip has area $y\,dx$, and the integral sums them from $x=a$ to $x=b$. Always sketch the region first. The picture fixes the limits, tells you whether the curve dips below the axis, and decides whether it is cleaner to integrate in $x$ or in $y$.

When the curve dips below the axis. Where $f(x)<0$ the strips have negative height, so $\int_a^b f(x)\,dx$ returns a negative value. Geometric area is always positive, so split the interval at every $x$-intercept and take the absolute value of each below-axis piece:

For $y=\sin x$ on $[0,2\pi]$, for instance, the plain integral is $0$ (the hump above cancels the hump below), but the area is $\int_0^{\pi}\sin x\,dx+\left|\int_{\pi}^{2\pi}\sin x\,dx\right|=2+2=4$.

Area with respect to the $y$-axis. When a region is bounded on the left by the $y$-axis, or is described more naturally as $x=g(y)$, use horizontal strips of height $dy$ and width $x$:

A sideways parabola such as $y^2=4ax$ or the part of a circle to the right of the $y$-axis is usually easiest this way.

Standard curves and their areas

A handful of standard regions recur throughout the exercise. Memorise the shapes; re-derive the formulae by integration.

Exploiting symmetry. Most standard curves are symmetric about an axis. For the circle $x^2+y^2=a^2$, the upper half is $y=\sqrt{a^2-x^2}$, so the whole area is

Working one quadrant and multiplying by $4$ (or one half and doubling) keeps the limits simple and avoids sign trouble. The two standard integrals you will lean on are $\int\sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+C$ and $\int\sqrt{x^2-a^2}\,dx$, both from the previous chapter.

Area bounded by a curve and a line

When one boundary of the region is a straight line (for example, find the area of $y^2=4ax$ cut off by its latus rectum $x=a$, or the area of the circle to one side of a chord), the method is unchanged: identify the limits from where the line meets the curve, then integrate $y\,dx$ (or $x\,dy$) over that span. The line simply supplies one of the limits or one of the ordinates.

Deeper Insight — the integral is a strip-summing machine, sign and all. Everything in this topic follows from one picture: a thin vertical strip of area $y\,dx$. If you keep that picture in mind, three things stop being separate rules. First, area below the axis comes out negative simply because the strip height $y$ is negative there — so you take $|\cdot|$, not because of a special formula. Second, integrating along $y$ is the very same idea turned $90^\circ$: strips of area $x\,dy$. Third, symmetry works because identical strips on either side of an axis contribute equally, so you may integrate a convenient quarter and scale up. Treat the limits as "where does the region start and stop", read them straight off your sketch, and the integration itself is routine.