Application of Integrals

🎯 Key Points

• Area is always non-negative, but the definite integral can be negative when the curve dips below the x-axis — take the ABSOLUTE VALUE of any below-axis piece before adding
• If a curve crosses the x-axis (or two curves cross each other) within the interval, SPLIT at the crossing point(s) first — never integrate straight through a sign change
• Area between curves = ∫(upper−lower)dx, always subtracting lower from upper so the integrand stays non-negative
• Standard enclosed areas: circle x²+y²=a² → πa²; ellipse x²/a²+y²/b²=1 → πab (circle is the special case a=b); these come from the SAME symmetry technique (4× first-quadrant area)

Application of Integrals

Definite integrals give a precise tool for computing areas of regions bounded by curves, something that simple geometric formulas cannot handle for curved boundaries.

Area Under a Simple Curve

If y = f(x) is continuous and non-negative on [a, b], the area bounded by the curve, the x-axis, and the lines x = a and x = b is:

Area = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx

Similarly, the area bounded by a curve x = g(y), the y-axis, and the lines y = c and y = d is Area = ∫꜀ᵅ x dy.

Sign Convention for Area

• A definite integral can be negative, but area is always taken as a non-negative quantity.
• If the curve lies below the x-axis over [a, b] (i.e. f(x) < 0), the definite integral ∫ₐᵇ f(x) dx comes out negative; the actual area is the absolute value, |∫ₐᵇ f(x) dx|.
• If a curve crosses the x-axis within the interval, split the interval at the crossing point(s), compute the integral on each piece separately, take the absolute value of any negative piece, and add the pieces together. Never integrate straight across a sign change and rely on cancellation.

Area Between Two Curves

If f(x) ≥ g(x) for all x in [a, b], the area enclosed between the two curves and the lines x = a, x = b is:

Area = ∫ₐᵇ [f(x) - g(x)] dx

Always subtract the lower curve from the upper curve so the integrand stays non-negative on the interval. If the curves cross inside [a, b], split at the intersection point(s) and adjust which function is "upper" in each sub-interval.

Step by Step Approach

• Sketch the curves (even roughly) to see which curve is on top/bottom or left/right, and where they intersect.
• Find points of intersection by solving the equations simultaneously, these become the limits of integration.
• Decide whether to integrate with respect to x or y, whichever gives a simpler integrand.
• Set up the integral as (upper - lower) or (right - left), watch the sign convention, then evaluate using the Fundamental Theorem of Calculus.

Area of a Circle

For the circle x² + y² = a², the area of the full circle using integration (by symmetry, 4 times the first-quadrant area) is:

Area = 4∫₀ᵃ √(a² - x²) dx = 4 × (πa²/4) = πa²

Area of a Parabola

For the parabola y² = 4ax (a > 0) and its latus rectum x = a, the area of the upper-half region (bounded by the parabola, the x-axis, and x = a) is ∫₀ᵃ √(4ax) dx = (4/3)a². By symmetry about the x-axis, the total area enclosed between the parabola and the latus rectum (both halves together) is:

Area = 2 × (4/3)a² = (8/3)a²

Area of an Ellipse

For the ellipse x²/a² + y²/b² = 1, by symmetry the total area (4 times the first-quadrant area) is:

Area = 4∫₀ᵃ (b/a)√(a² - x²) dx = 4 × (πab/4) = πab

This generalizes the circle formula (when a = b = r, πab becomes πr²).

Worked Style Examples

• Area under y = x² from x = 0 to x = 2: ∫₀² x² dx = [x³/3]₀² = 8/3 sq units.
• Area between y = x and y = x² from x = 0 to x = 1: since x ≥ x² on [0,1], Area = ∫₀¹ (x - x²) dx = [x²/2 - x³/3]₀¹ = 1/2 - 1/3 = 1/6 sq units.

Area Bounded by a Curve and a Line

When a region is enclosed between a curve and a straight line (a chord), the limits of integration are the x-coordinates (or y-coordinates) of their points of intersection, which are NOT given directly and must be found first.

• Step 1: solve the curve and the line simultaneously to obtain the points of intersection.
• Step 2: decide which equation is the upper (or right) boundary over the region.
• Step 3: Area = ∫ (upper - lower) dx between the two intersection abscissae.
• Example: the line y = 3x meets the parabola y = x² at (0, 0) and (3, 9); since 3x ≥ x² on [0, 3], the enclosed area = ∫₀³ (3x - x²) dx = [3x²/2 - x³/3]₀³ = 27/2 - 9 = 9/2 sq units.

Area by Integrating Along the y-axis (Horizontal Strips)

When a boundary is more naturally written as x = g(y), it is easier to use horizontal strips of thickness dy and integrate with respect to y.

• Area = ∫꜀ᵈ x dy = ∫꜀ᵈ g(y) dy, where c and d are the y-limits of the region.
• For a region between two such curves, Area = ∫꜀ᵈ (x_right - x_left) dy.
• This is preferable when the "upper" curve changes across the region, or when a boundary is the y-axis or a sideways-opening parabola x = y²/4a — it avoids splitting the region into several x-intervals.
• Example: the area bounded by x = y², the y-axis, and the lines y = 0 and y = 3 is ∫₀³ y² dy = [y³/3]₀³ = 9 sq units.

🚀 JEE Advanced Edge

Choosing to integrate with respect to y instead of x: When the region's boundary is more naturally described as x=g(y) (e.g. a sideways-opening parabola, or when the "upper/lower" relationship between curves switches multiple times if viewed along x), integrating with respect to y (horizontal strips) instead of x (vertical strips) avoids splitting the region into multiple x-intervals — recognizing when to switch axes often turns a multi-step problem into a single integral.

Finding area enclosed by a curve and a chord/line using points of intersection: For the area between a parabola and a line (or any curve and a chord), the limits of integration are NOT given directly — they must be found by solving the curve and line equations simultaneously first, then those x-values (or y-values) become a and b in the area integral.

Worked problem: Find the area enclosed between the parabola y²=4x and the line y=x. Approach: Find intersections: substitute y=x into y²=4x → x²=4x → x²-4x=0 → x=0 or x=4 (giving points (0,0) and (4,4)). Integrating with respect to x, the parabola gives y=2√x (upper) and the line gives y=x (lower, since the parabola is above the line for 0<x<4 — check at x=1: parabola y=2, line y=1). Area = ∫₀⁴ (2√x - x)dx = [4x^(3/2)/3 - x²/2]₀⁴ = (4×8/3 - 8) - 0 = 32/3 - 8 = 8/3 sq units.

Worked Example: Area Between a Curve and the X-Axis

Find the area bounded by y = x² − 4 and the x-axis between x = −2 and x = 2.

Note: y = x² − 4 ≤ 0 on [−2, 2] (the parabola dips below the axis), so the area = |∫₋₂² (x²−4) dx|.

∫₋₂² (x²−4) dx = [x³/3 − 4x]₋₂² = (8/3 − 8) − (−8/3 + 8) = −16/3 − 16/3 = −32/3. Area = 32/3 sq units. When the curve is below the x-axis, take the absolute value of the integral — it's a very common sign error in exams.