Area under a Curve (Definite Integral)

If the graph runs above the x-axis on [a; b], the area is directly the definite integral: A = ∫ₐᵇ f(x) dx. In practice: find an antiderivative, evaluate at the limits, subtract. Example: the area under f(x) = x² between 0 and 2 is ∫₀² x² dx = [x³/3]₀² = 8/3 ≈ 2.67 area units. Before computing, always check that f really is nonnegative on the whole interval, quickest via its zeros. If the graph partly runs below the axis, you must split the interval at the zeros, otherwise signed areas offset each other and you get a value that is too small or even negative.

Then the integral gives a negative value, because it measures signed areas: positive above, negative below. For the geometric area you take the absolute value of the partial integral. If the graph runs partly above and partly below the axis, proceed in three steps: find the zeros in the interval, integrate piecewise between the zeros, add the absolute values of the partial results. Warning example: ∫₋₁¹ x³ dx = 0, although the curve encloses genuine areas; the left part (−1/4) and right part (+1/4) merely cancel arithmetically. The actual area is |−1/4| + |1/4| = 1/2 area units. Exams love to test exactly this difference between integral value and area.

Integrate the difference function: A = ∫ₐᵇ (f(x) − g(x)) dx, where f runs above g on the interval. The limits are usually the intersection points, found from f(x) = g(x). Example: f(x) = x and g(x) = x² intersect at 0 and 1; there x ≥ x², so A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6 area units. Convenient: it does not matter whether the graphs lie above or below the x-axis, the difference compensates automatically. If the curves intersect inside the interval, you must split at the intersections, because f − g changes sign there, and add the absolute values of the partial integrals.

Because the definite integral balances signed areas and does not add geometric areas. Every region above the x-axis counts positive, every one below negative; the integral is the sum of these signed contributions. In point-symmetric situations like ∫₋₂² x³ dx the two contributions cancel exactly, the integral is 0, although genuine pieces of area lie on both sides. This balance logic is not a defect but often exactly what is wanted: in a v-t diagram a negative sign means backward motion, and the integral gives the net displacement rather than the distance travelled. If you want the area instead, integrate between the zeros separately and add the absolute values.

Whenever a product of two quantities is summed over an interval. In physics the area in the velocity-time diagram is the distance covered, the area in the force-displacement diagram the work done, and the area under the power curve the energy. In stochastics, probabilities of continuous random variables are areas under the density function, whose total area is always 1. Economics computes consumer and producer surplus as areas between price and demand curves. Averages are area logic too: the mean value of a function is its area divided by the interval length. Whoever masters the area interpretation can interpret integrals in almost any applied context.