What this chapter is about
This chapter shows how definite integrals move beyond pure calculation to solve a geometric problem: finding the area of a region bounded by curves. In Class 11 and earlier in Class 12, you learned to evaluate integrals; now you apply that skill to measure plane areas that simple length × breadth cannot handle.
The central idea is that a definite integral ∫[a to b] f(x) dx gives the signed area between the curve y = f(x), the x-axis, and the vertical lines x = a and x = b. When a region is bounded by two curves, you find the area between them by integrating the difference of their ordinates. This connects algebra, coordinate geometry and calculus in one powerful technique.
After working through this chapter, you should be able to sketch the region whose area is required, set up the correct integral (or sum of integrals), handle curves above and below the axis, and compute the final numerical value in square units.
Key ideas
- The definite integral ∫[a to b] f(x) dx represents the net signed area: positive where f(x) ≥ 0, negative where f(x) < 0.
- To find the actual (unsigned) area when a curve dips below the x-axis, split the interval at the zeros and take the absolute value of each part.
- Area between two curves y = f(x) and y = g(x) from x = a to x = b, with f(x) ≥ g(x) throughout, is ∫[a to b] [f(x) − g(x)] dx.
- When curves cross inside the interval, locate the intersection points and split the integral accordingly, always subtracting the lower curve from the upper in each sub-interval.
- Sometimes it is simpler to integrate with respect to y: area = ∫[c to d] [right curve − left curve] dy.
- Sketching the region first prevents sign errors and helps identify limits of integration.
Formulas and facts to remember
- Situation: Area under y = f(x) above the x-axis, from x = a to b · Formula: A = ∫[a to b] f(x) dx · Meaning: Sum of thin vertical strips of width dx and height f(x).
- Situation: Area when curve lies below the x-axis · Formula: A = −∫[a to b] f(x) dx or ∫[a to b] |f(x)| dx · Meaning: The integral itself is negative; take magnitude for area.
- Situation: Area between y = f(x) (upper) and y = g(x) (lower) · Formula: A = ∫[a to b] [f(x) − g(x)] dx · Meaning: Difference of ordinates, integrated over the common x-range.
- Situation: Area using horizontal strips · Formula: A = ∫[c to d] [x_right(y) − x_left(y)] dy · Meaning: Useful when the boundary is easier to express as x in terms of y.
Standard results worth recalling:
- Area of a circle x² + y² = r² is πr² (can be derived by integrating y = √(r² − x²) from −r to r and doubling).
- Area of an ellipse x²/a² + y²/b² = 1 is πab.
Worked examples
### Example 1 Find the area enclosed between the parabola y = x² and the line y = 2x.
Step 1 – Find intersection points. Set x² = 2x ⟹ x² − 2x = 0 ⟹ x(x − 2) = 0. So x = 0 and x = 2.
Step 2 – Identify upper and lower curves. For 0 < x < 2, check x = 1: line gives y = 2, parabola gives y = 1. The line y = 2x is above the parabola y = x².
Step 3 – Set up and evaluate the integral. A = ∫[0 to 2] (2x − x²) dx = [x² − x³/3] from 0 to 2 = (4 − 8/3) − 0 = 12/3 − 8/3 = 4/3 square units.
### Example 2 Find the area of the region bounded by y = sin x, the x-axis, x = 0 and x = π.
Step 1 – Note the sign of y. For 0 ≤ x ≤ π, sin x ≥ 0, so the curve lies on or above the x-axis.
Step 2 – Integrate directly. A = ∫[0 to π] sin x dx = [−cos x] from 0 to π = (−cos π) − (−cos 0) = (−(−1)) − (−1) = 1 + 1 = 2 square units.
### Example 3 Find the area enclosed in the first quadrant by the circle x² + y² = 16 and the line x + y = 4.
Step 1 – Find intersection in the first quadrant. Circle: radius 4, centred at origin. Line: x + y = 4 ⟹ y = 4 − x. Substitute into circle: x² + (4 − x)² = 16 x² + 16 − 8x + x² = 16 ⟹ 2x² − 8x = 0 ⟹ 2x(x − 4) = 0. So x = 0 (then y = 4) and x = 4 (then y = 0).
Step 2 – Identify upper curve. Between x = 0 and x = 4 in the first quadrant, the circle's upper half is y = √(16 − x²), and it lies above the chord y = 4 − x.
Step 3 – Set up the integral. Area between them = ∫[0 to 4] [√(16 − x²) − (4 − x)] dx.
Step 4 – Evaluate each part. ∫ √(16 − x²) dx from 0 to 4 uses the standard result: (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a). Here a = 4. At x = 4: (4/2)·0 + (16/2)·(π/2) = 4π. At x = 0: 0 + 0 = 0. So the first integral = 4π.
∫ (4 − x) dx from 0 to 4 = [4x − x²/2] from 0 to 4 = 16 − 8 = 8.
Area = 4π − 8 square units (approximately 4.57 square units).
Common mistakes
- Forgetting to take the absolute value when the curve goes below the x-axis → split at zeros and add magnitudes.
- Subtracting in the wrong order (lower minus upper) and getting a negative area → always upper curve minus lower curve.
- Using wrong limits because the sketch was skipped → always draw a rough diagram first.
- Confusing the roles of x and y when curves are easier to express as x = h(y) → switch to horizontal strips and integrate with respect to y.
- Applying the area formula for a full circle or ellipse when only a segment or sector is asked → integrate only over the required region.
Quick revision
- Definite integral = signed area; take absolute value for actual area below the axis.
- Area between curves = ∫ (upper − lower) dx over the interval where they enclose a region.
- Always find intersection points first; they become limits of integration.
- Sketch the region to decide which function is on top.
- Use ∫ dy (horizontal strips) when boundaries are simpler in terms of y.