Area under Simple Curves

From Definite Integrals to Areas

Elementary geometry hands us area formulas for triangles, rectangles, trapeziums and circles — but nothing for regions bounded by curves. Integral calculus fills that gap. The previous chapter established that a definite integral is a number computed via the Fundamental Theorem of Calculus; this chapter reads that number as an area.

The elementary strip

To find the area bounded by the curve $y = f(x)$, the $x$-axis and the ordinates $x = a$ and $x = b$, imagine the region sliced into a large number of very thin vertical strips. An arbitrary strip has height $y$ and width $dx$, so its area is

$$dA = y\,dx, \quad \text{where } y = f(x)$$

This elementary area sits at an arbitrary position $x$ between $a$ and $b$. Adding up all the strips across the region:

$$A = \int_a^b dA = \int_a^b y\,dx = \int_a^b f(x)\,dx$$

Horizontal strips — area against the $y$-axis

If the region is bounded by the curve $x = g(y)$, the $y$-axis and the lines $y = c$, $y = d$, slice it into thin horizontal strips of length $x$ and width $dy$:

$$A = \int_c^d x\,dy = \int_c^d g(y)\,dy$$

Key Point: The strip is the whole method. Decide which axis the region leans on, write the area of ONE strip ($y\,dx$ or $x\,dy$), then integrate between the limits. Every problem in this chapter is those three moves plus a sketch.

Choosing the strip direction

1. Boundaries of the form $x = a$, $x = b$ with a curve $y = f(x)$ on top → vertical strips, integrate in $x$.
2. Boundaries of the form $y = c$, $y = d$ with a curve $x = g(y)$ on the right → horizontal strips, integrate in $y$.
3. Many regions can be done both ways (the circle in the next section is NCERT's own demonstration) — pick the direction in which the curve solves cleanly for the strip length.

[JEE Tip] In JEE Main, choosing $dy$ strips instead of $dx$ often converts a two-piece integral into a single clean one — for example, the region bounded by $y^2 = 4x$, the $y$-axis and $y = 3$ is a one-line $\int_0^3 \frac{y^2}{4}\,dy$ with horizontal strips, but needs a subtraction setup with vertical ones.

The Sign Rule — Curves Below the Axis

If the curve lies below the $x$-axis between $x = a$ and $x = b$ (that is, $f(x) < 0$ there), each strip height $y$ is negative, so the integral comes out negative. But area is a magnitude — only the numerical value counts. So for a region entirely below the axis:

$$A = \left|\int_a^b f(x)\,dx\right|$$

Mixed regions — the $|A_1| + A_2$ rule

Generally a curve may run partly above and partly below the axis. If the piece below contributes $A_1 < 0$ and the piece above contributes $A_2 > 0$, the bounded area is

$$A = |A_1| + A_2$$

The working procedure:

1. Sketch the curve and mark where it crosses the $x$-axis inside $[a, b]$ (solve $f(x) = 0$).
2. Split the integral at every crossing point.
3. Integrate each piece separately, take the absolute value of each, and add.

Key Point: $\int_a^b f(x)\,dx$ computes the signed sum $A_1 + A_2$, not the area. Blind integration across a sign change silently cancels area against area — the single most common error of this chapter. Split first, then integrate.

Contrast worth memorising: $\int_0^{2\pi}\sin x\,dx = 0$ (signed), but the area between $y = \sin x$ and the $x$-axis over $[0, 2\pi]$ is $4$ (two arches of area $2$ each).

[JEE Tip] JEE sets exactly this trap with odd powers: the area bounded by $y = x^3$, the $x$-axis, $x = -2$ and $x = 1$ is $\left|\int_{-2}^0 x^3 dx\right| + \int_0^1 x^3 dx = 4 + \frac14 = \frac{17}{4}$, while the unsplit integral gives $-\frac{15}{4}$ — and both wrong values appear among the options.

Solved Examples

Example 1: A triangle, two ways

Find the area bounded by the line $y = 2x$, the $x$-axis and the ordinates $x = 0$ and $x = 3$.

Solution:

1. Set up vertical strips: $A = \int_0^3 y\,dx = \int_0^3 2x\,dx$.
2. Evaluate: $\left[x^2\right]_0^3 = 9$.
3. Geometric check: the region is a triangle with base $3$ and height $6$: $\frac12\cdot3\cdot6 = 9$. ✓

Final Answer: $9$ square units.

Example 2: Area under a parabola

Find the area under $y = x^2$ from $x = 0$ to $x = 2$.

Solution:

1. Set up: the parabola lies above the axis, so $A = \int_0^2 x^2\,dx$.
2. Evaluate: $\frac{x^3}{3}\Big|_0^2 = \frac83$.

Final Answer: $\dfrac{8}{3}$ square units — one third of the $2 \times 4$ bounding rectangle, the classic parabola fraction.

Example 3: A root curve

Find the area bounded by $y = \sqrt x$, the $x$-axis and $x = 4$.

Solution:

1. Set up: $A = \int_0^4 \sqrt x\,dx$.
2. Evaluate: $\frac{2}{3}x^{3/2}\Big|_0^4 = \frac23\cdot 8 = \frac{16}{3}$.

Final Answer: $\dfrac{16}{3}$ square units.

Example 4: Horizontal strips

Find the area of the region bounded by the curve $x = y^2$, the $y$-axis and the lines $y = 1$, $y = 3$.

Solution:

1. Choose the strip: the region leans on the $y$-axis with $x$ given in terms of $y$ — horizontal strips: $A = \int_1^3 x\,dy = \int_1^3 y^2\,dy$.
2. Evaluate: $\frac{y^3}{3}\Big|_1^3 = \frac{27 - 1}{3} = \frac{26}{3}$.

Final Answer: $\dfrac{26}{3}$ square units.

Takeaway: No conversion to $y = \sqrt x$ was needed — when the curve arrives as $x = g(y)$, ride the $y$-axis and integrate in $y$ directly.

Example 5: An exponential region

Find the area bounded by $y = e^x$, the $x$-axis and the ordinates $x = 0$, $x = 1$.

Solution:

1. Set up: $e^x > 0$ everywhere, so no sign worries: $A = \int_0^1 e^x dx$.
2. Evaluate: $\left[e^x\right]_0^1 = e - 1$.

Final Answer: $e - 1$ square units (about $1.72$).

Example 6: The hyperbola arm

Find the area bounded by $y = \frac1x$, the $x$-axis and the ordinates $x = 1$, $x = 4$.

Solution:

1. Set up: on $[1, 4]$, $\frac1x > 0$: $A = \int_1^4\frac{dx}{x}$.
2. Evaluate: $\left[\log x\right]_1^4 = \log 4$.

Final Answer: $\log 4 = 2\log 2$ square units.

Takeaway: Areas under $\frac1x$ are logarithms — this is exactly why $\log$ appears in nature as accumulated growth.

Example 7: One arch of sine

Find the area bounded by $y = \sin x$ and the $x$-axis between $x = 0$ and $x = \pi$.

Solution:

1. Check the sign: $\sin x \geq 0$ on all of $[0, \pi]$ — a single arch above the axis.
2. Evaluate: $\int_0^{\pi}\sin x\,dx = \left[-\cos x\right]_0^{\pi} = 1 + 1 = 2$.

Final Answer: $2$ square units — the famous "area of one sine arch".

Example 8: Below the axis

Find the area bounded by $y = x^3$, the $x$-axis and the ordinates $x = -2$, $x = 0$.

Solution:

1. Check the sign: $x^3 \leq 0$ on $[-2, 0]$ — the region lies entirely below the axis.
2. Integrate: $\int_{-2}^{0} x^3\,dx = \frac{x^4}{4}\Big|_{-2}^{0} = 0 - 4 = -4$.
3. Take the absolute value: area $= |-4| = 4$.

Final Answer: $4$ square units.

Takeaway: A negative answer from an area integral is a message, not a mistake — it says the region is below the axis. Report its absolute value, and split first whenever the curve crosses the axis inside the interval.