Positive / negative area concept, symmetry properties, and key results that save time in exams.
Class 12 · Mathematics · CBSE & JEEA definite integral gives a signed value. When the curve is above the x-axis, the integral is positive. When below, the integral is negative. But area is always positive.
If \(f(x) \geq 0\) on \([a,b]\), the curve lies above the x-axis.
\(\displaystyle\int_a^b f(x)\,dx > 0\)
Area = this value directly.
If \(f(x) \leq 0\) on \([a,b]\), the curve lies below the x-axis.
\(\displaystyle\int_a^b f(x)\,dx < 0\)
Area = \(\left|\displaystyle\int_a^b f(x)\,dx\right|\)
Zeros: \(x^3-x=0 \Rightarrow x(x^2-1)=0 \Rightarrow x = -1,0,1\). Crosses at \(x=0\).
On \([-1,0]\): \(x^3-x = x(x^2-1) = x(x-1)(x+1)\). At \(x=-0.5\): \((-0.5)(0.75)\approx 0.375 > 0\). Positive.
On \([0,1]\): At \(x=0.5\): \(0.125-0.5 < 0\). Negative.
\(A_1 = \displaystyle\int_{-1}^0(x^3-x)\,dx = \left[\dfrac{x^4}{4}-\dfrac{x^2}{2}\right]_{-1}^0 = 0-\left(\dfrac{1}{4}-\dfrac{1}{2}\right) = \dfrac{1}{4}\)
\(A_2 = \left|\displaystyle\int_0^1(x^3-x)\,dx\right| = \left|\left[\dfrac{x^4}{4}-\dfrac{x^2}{2}\right]_0^1\right| = \left|\dfrac{1}{4}-\dfrac{1}{2}\right| = \dfrac{1}{4}\)
Total Area = \(\dfrac{1}{4}+\dfrac{1}{4} = \mathbf{\dfrac{1}{2}}\) sq. units
Note: if you blindly computed \(\int_{-1}^1(x^3-x)dx\) you'd get 0 — completely wrong!
Symmetry is the most powerful tool to reduce computation. Recognise it and halve (or quarter) your work.
Geometric meaning: The region is symmetric about the y-axis. Left half mirrors right half.
Examples: \(y = x^2\), \(y = \cos x\), \(y = |x|\), \(x^2+y^2=r^2\) (circle).
Geometric meaning: The positive area and negative area cancel out exactly.
For actual area: \(A = 2\displaystyle\int_0^a |f(x)|\,dx = 2\displaystyle\int_0^a f(x)\,dx\) (if \(f \geq 0\) on \([0,a]\)).
Examples: \(y = x^3\), \(y = \sin x\), \(y = x\).
Always compute one quarter and multiply by 4. This is the standard method in board exams.
For \(y^2 = 4ax\): use \(y = 2\sqrt{ax}\) (upper half) and multiply by 2.
| # | Property | Formula | Use in Area |
|---|---|---|---|
| P1 | Reversal of limits | \(\int_a^b f\,dx = -\int_b^a f\,dx\) | Fixing negative results from wrong limit order |
| P2 | Additivity | \(\int_a^b = \int_a^c + \int_c^b\) | Splitting at zero-crossing points |
| P3 | Even function | \(\int_{-a}^a f = 2\int_0^a f\) (if even) | Halving computation for symmetric regions |
| P4 | Odd function | \(\int_{-a}^a f = 0\) (if odd) | Quickly evaluating areas — but get actual area by doubling one side |
| P5 | Periodic function | \(\int_0^{nT} f = n\int_0^T f\) | Area under \(n\) cycles of \(\sin x\), \(\cos x\) |
| P6 | Modulus | \(\int_a^b|f|\,dx \geq |\int_a^b f\,dx|\) | Area ≥ absolute value of integral |
Computing \(\int_{-a}^a \sin x\,dx = 0\) and stating area = 0. Area is 2, not 0! (odd function integral = 0, but area ≠ 0).
Not splitting the integral when the curve crosses the x-axis. Always find zeros first.
For area between curves, subtracting in the wrong order. Always: upper − lower or right − left.
Using area = \(\int_a^b f\,dx\) directly when \(f(x)\) can be negative. Take absolute value when needed.
For circle area: computing \(\int_{-r}^r\sqrt{r^2-x^2}\,dx\) instead of \(4\int_0^r...\,dx\). Both are correct but the latter is faster.
Forgetting to double the area when using symmetry. If you integrate one half, always multiply by 2 (or 4 for circular symmetry).