Area Bounded by a Curve, Lines & Axes

For a function $y = f(x)$ and an interval $[a, b]$, the definite integral measures the signed area between the curve and the x-axis:

• If $f(x) \ge 0$ on $[a, b]$ (curve above the axis), the integral equals the geometric area, which is always positive.
• If $f(x) \le 0$ (curve below the axis), the integral is negative; the geometric area is its absolute value.
• The two vertical lines $x = a$ and $x = b$ are the limits — read them off as where the region starts and ends. The x-axis itself is $y = 0$.

To find the area bounded by $y = f(x)$, the x-axis, and the lines $x = a$, $x = b$:

• Set up: $A = \int_a^b f(x)\,dx$, provided $f \ge 0$ on $[a, b]$.
• Recognise known shapes instead of integrating when you can:
• $y = \sqrt{r^2 - x^2}$ is the upper semicircle of radius $r$; its area is $\tfrac{1}{2}\pi r^2$.
• A region cut by straight lines is a triangle or rectangle — use $\tfrac{1}{2}\,\text{base}\times\text{height}$ or length×breadth.
• For a trig boundary like $y = \cos x$ on a subinterval, just integrate: $\int \cos x\,dx = \sin x$, $\int \sin x\,dx = -\cos x$.

Geometric area never cancels. When the curve crosses the axis or the region is symmetric:

• Split at every crossing. If $f$ changes sign at $x = c$ inside $[a, b]$, then

• Use symmetry as a shortcut. For a region symmetric about the y-axis (or about a point), area on one side equals the other: $A = 2 \times (\text{area of one half})$. A loop of $y = c\sin x$ runs over one half-period.
• A function like $f(x) = x|x|$ equals $x^2$ for $x > 0$ and $-x^2$ for $x < 0$ — equal areas on each side, so total area $= 2\int_0^{\,k} x^2\,dx$.

Modulus equations unfold into straight-line pieces, fencing off a polygon:

• $|x| + |y| = 1$ is a square (a tilted diamond) with vertices $(\pm 1, 0)$ and $(0, \pm 1)$; diagonal $2$, area $2$.
• $|x| \le p$ and $|y| \le q$ is a rectangle of width $2p$ and height $2q$: area $4pq$.
• $x = |y|$ is a sideways V; with a vertical line $x = c$ it closes a triangle of base $2c$ (the vertical side) and height $c$.

Method: sketch, find the corner points, then apply length×breadth (rectangle) or $\tfrac{1}{2}\,\text{base}\times\text{height}$ (triangle).

For the right-opening parabola $y^2 = 4ax$:

• The latus rectum is the vertical chord through the focus, the line $x = a$.
• The upper boundary is $y = \sqrt{4ax}$; the region is symmetric about the x-axis, so

• Read the limit off the equation: for $y^2 = 4kx$ the latus rectum is $x = k$; for $y^2 = x$ write it as $y^2 = 4(\tfrac14)x$, so $a = \tfrac14$ and the limit is $x = \tfrac14$.

For a piecewise-constant boundary such as the greatest-integer function $y = [x]$:

• On any interval where $[x]$ holds a single value $n$, the region is a rectangle of height $|n|$ and width = interval length.
• $[x] = n$ for $n \le x < n+1$; e.g. for $x \in [-1.8, -1.5]$, every value lies in $[-2, -1)$, so $[x] = -2$ throughout.
• Area $= |n| \times (\text{width})$. For a multi-step interval, sum the rectangles.

When a chord (here a line such as $y = x$) cuts a circle into two regions $A_1$ (major) and $A_2$ (minor):

• The minor segment $A_2 = (\text{sector area}) - (\text{triangle area})$; set it up as a definite integral between the chord and the arc.
• The major segment $A_1 = (\text{full circle area}) - A_2 = \pi r^2 - A_2$.
• For the unit-radius circle $(x-1)^2 + y^2 = 1$ cut by $y = x$ (chord from $(0,0)$ to $(1,1)$): $A_2 = \tfrac{\pi - 2}{4}$ and $A_1 = \pi - A_2 = \tfrac{3\pi + 2}{4}.$