Lines, Modulus and Trigonometric Curves

Sketch First, Integrate Second

The regions in this section all share one feature: the curve crosses the $x$-axis inside the interval, or arrives wrapped in a modulus. Blind integration fails on every one of them. The reliable workflow:

1. Sketch the curve — even a rough sign-correct sketch.
2. Find the $x$-intercepts inside the interval (solve $y = 0$).
3. Split the integral at each intercept.
4. Integrate each piece, take absolute values, add.

The model case: a line crossing the axis

For the area bounded by $y = 3x + 2$, the $x$-axis and the ordinates $x = -1$, $x = 1$: the line meets the axis at $x = -\frac23$, lying below for $-1 \leq x < -\frac23$ and above for $-\frac23 < x \leq 1$.

$$A = \left|\int_{-1}^{-2/3}(3x + 2)\,dx\right| + \int_{-2/3}^{1}(3x + 2)\,dx = \left|-\frac{1}{6}\right| + \frac{25}{6} = \frac{13}{3}$$

Skipping the split gives $\int_{-1}^{1}(3x+2)dx = 4$ — a wrong area, because the piece below the axis entered with the wrong sign.

Key Point: An $x$-intercept inside the interval is a mandatory checkpoint. Two clean sub-integrals beat one wrong shortcut — and both boards and JEE deliberately choose intervals that straddle an intercept.

[JEE Tip] For straight lines the two pieces are triangles — check each integral against $\frac12\times\text{base}\times\text{height}$. Here the left triangle has base $\frac13$ and height $1$: area $\frac16$ ✓, and the right one has base $\frac53$ and height $5$: area $\frac{25}{6}$ ✓. Free verification, zero extra time.

Trigonometric Arches and Modulus Curves

Full-period trig areas

Over $[0, 2\pi]$, $y = \cos x$ is positive on $\left[0, \frac{\pi}{2}\right]$, negative on $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$, positive again on $\left[\frac{3\pi}{2}, 2\pi\right]$ — three regions:

$$A = \int_0^{\pi/2}\cos x\,dx + \left|\int_{\pi/2}^{3\pi/2}\cos x\,dx\right| + \int_{3\pi/2}^{2\pi}\cos x\,dx = 1 + 2 + 1 = 4$$

The same argument gives area $4$ for $y = \sin x$ over $[0, 2\pi]$ (two arches of $2$ each). The signed integrals are $0$ in both cases — maximal cancellation, maximal trap.

Arch arithmetic worth memorising: every half-arch of $\sin x$ or $\cos x$ (a quarter period) has area $1$; every full arch has area $2$; a full period has area $4$.

Modulus curves

1. $y = |x + 3|$: a V with vertex at $(-3, 0)$. On $[-6, 0]$ it makes two congruent triangles (base $3$, height $3$ each): $\int_{-6}^{0}|x + 3|\,dx = \frac92 + \frac92 = 9$. Working rule: replace $|x+3|$ by $-(x+3)$ left of $-3$ and $(x+3)$ right of it.
2. $y = x|x|$: equals $x^2$ for $x \geq 0$ and $-x^2$ for $x < 0$ — an odd S-shaped curve through the origin. The area bounded with the $x$-axis on $[-1, 1]$ is $\left|\int_{-1}^0(-x^2)dx\right| + \int_0^1 x^2dx = \frac13 + \frac13 = \frac23$ (while the signed integral is $0$, since the function is odd).
3. Power curves with negative stretches: $y = x^3$ on $[-2, 1]$ gives area $\left|\int_{-2}^{0}x^3dx\right| + \int_0^1 x^3dx = 4 + \frac14 = \frac{17}{4}$.

Key Point: A modulus is an instruction to split at the corner or sign change and rewrite each branch without the modulus. After that it's ordinary integration — the modulus never survives into the anti-derivative.

[JEE Tip] Symmetry still helps with areas — but with absolute values: for odd $f$, the area on $[-a, a]$ is $2\int_0^a|f|$, not $0$. Odd symmetry kills signed integrals, and doubles areas.

Solved Examples

Example 1: The line that crosses

Find the area of the region bounded by the line $y = 3x + 2$, the $x$-axis and the ordinates $x = -1$ and $x = 1$.

Solution:

1. Find the intercept: $3x + 2 = 0$ at $x = -\frac23$, which lies inside $[-1, 1]$ — split there.
2. Left piece (below the axis): $\int_{-1}^{-2/3}(3x+2)dx = \left[\frac{3x^2}{2} + 2x\right]_{-1}^{-2/3} = -\frac23 - \left(-\frac12\right) = -\frac16$; take $\left|-\frac16\right| = \frac16$.
3. Right piece (above): $\int_{-2/3}^{1}(3x+2)dx = \frac72 - \left(-\frac23\right) = \frac{25}{6}$.
4. Add: $\frac16 + \frac{25}{6} = \frac{26}{6}$.

Final Answer: $\dfrac{13}{3}$ square units.

Example 2: Cosine over a full period

Find the area bounded by $y = \cos x$ and the $x$-axis between $x = 0$ and $x = 2\pi$.

Solution:

1. Sign chart: positive on $\left[0, \frac{\pi}{2}\right]$, negative on $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$, positive on $\left[\frac{3\pi}{2}, 2\pi\right]$.
2. Three pieces: $[\sin x]_0^{\pi/2} = 1$; $[\sin x]_{\pi/2}^{3\pi/2} = -2$, absolute value $2$; $[\sin x]_{3\pi/2}^{2\pi} = 1$.
3. Add: $1 + 2 + 1 = 4$.

Final Answer: $4$ square units.

Example 3: A plain power curve

Find the area under $y = x^2$ between $x = 1$, $x = 2$ and the $x$-axis.

Solution:

1. No sign worries: $x^2 > 0$ on $[1, 2]$.
2. Evaluate: $\int_1^2 x^2dx = \frac{x^3}{3}\Big|_1^2 = \frac{8 - 1}{3}$.

Final Answer: $\dfrac{7}{3}$ square units.

Example 4: A higher power

Find the area under $y = x^4$, between $x = 1$, $x = 5$ and the $x$-axis.

Solution:

1. Set up: $A = \int_1^5 x^4dx = \frac{x^5}{5}\Big|_1^5$.
2. Evaluate: $\frac{3125 - 1}{5} = \frac{3124}{5}$.

Final Answer: $\dfrac{3124}{5}$ square units ($624.8$).

Example 5: A modulus V-curve

Sketch $y = |x + 3|$ and evaluate $\displaystyle\int_{-6}^{0}|x + 3|\,dx$.

Solution:

1. Sketch: a V with vertex at $(-3, 0)$; on $[-6, -3]$, $|x+3| = -(x+3)$; on $[-3, 0]$, $|x+3| = x+3$.
2. Left piece: $\int_{-6}^{-3}-(x+3)dx = \left[-\frac{(x+3)^2}{2}\right]_{-6}^{-3} = 0 + \frac92 = \frac92$.
3. Right piece: $\int_{-3}^{0}(x+3)dx = \frac{(x+3)^2}{2}\Big|_{-3}^{0} = \frac92$.
4. Add: $\frac92 + \frac92 = 9$. (Check: two triangles, each $\frac12\cdot3\cdot3 = \frac92$. ✓)

Final Answer: $9$ square units.

Example 6: Sine over a full period

Find the area bounded by $y = \sin x$ and the $x$-axis between $x = 0$ and $x = 2\pi$.

Solution:

1. Sign chart: positive on $[0, \pi]$, negative on $[\pi, 2\pi]$ — two arches.
2. Pieces: $\int_0^{\pi}\sin x\,dx = 2$; $\left|\int_{\pi}^{2\pi}\sin x\,dx\right| = |-2| = 2$.
3. Add: $2 + 2 = 4$.

Final Answer: $4$ square units — while the signed integral over the same interval is $0$.

Example 7: An odd cube across the origin

Find the area bounded by $y = x^3$, the $x$-axis and the ordinates $x = -2$ and $x = 1$.

Solution:

1. Intercept inside: $x^3 = 0$ at $x = 0$ — split there.
2. Left piece: $\int_{-2}^{0}x^3dx = -4$; absolute value $4$.
3. Right piece: $\int_0^1 x^3dx = \frac14$.
4. Add: $4 + \frac14 = \frac{17}{4}$.

Final Answer: $\dfrac{17}{4}$ square units. (The unsplit integral gives $-\frac{15}{4}$ — a listed wrong option in the exercise MCQ.)

Example 8: The signature curve $y = x|x|$

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

Solution:

1. Unwrap the modulus: $y = x^2$ for $x \geq 0$ and $y = -x^2$ for $x < 0$ (an odd curve, below the axis on the left).
2. Left piece: $\int_{-1}^{0}(-x^2)\,dx = \left[-\frac{x^3}{3}\right]_{-1}^{0} = 0 - \frac13 = -\frac13$; take the absolute value: $\frac13$.
3. Right piece: $\int_0^1 x^2 dx = \frac13$.
4. Add: $\frac13 + \frac13 = \frac23$.

Final Answer: $\dfrac{2}{3}$ square units.

Takeaway: Odd curves bound equal areas on both sides of the origin — the signed integral cancels to $0$, the area doubles to $2\times$ one side. Know which of the two the question asks for.