Area Under a Curve Using Definite Integral

1. Identify the curve y = f(x). 2. Identify the lower and upper x-limits. 3. Check whether the curve is above the x-axis on the interval. 4. Write area as ∫_a^b f(x) dx or split if the sign changes. 5. Integrate, substitute limits, simplify, and write square units.

Draw x-axis and y-axis, sketch y = f(x), shade the region between the curve and x-axis from x = a to x = b, and label vertical boundaries x = a and x = b.

Using ∫_-2^2 x^2 - 4 dx for y = 4 - x^2 gives a negative value, because the integrand has been reversed relative to the shaded region.

The area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is found by adding thin vertical strips from x = a to x = b. If f(x) is non-negative on [a, b], the area is ∫_a^b f(x) dx.

The area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is found by adding thin vertical strips from x = a to x = b. If f(x) is non-negative on [a, b], the area is ∫_a^b f(x) dx. Key rule: Area = ∫_a^b y dx = ∫_a^b f(x) dx when the region is bounded by y = f(x), x-axis, x = a, x = b, and f(x) ≥ 0 on [a, b]. If f(x) ≤ 0, actual area = -∫_a^b f(x) dx. If f(x) changes sign, split the interval at zeros of f(x). The area bounded by y = x^2, the x-axis, x = 0 and x = 2 is ∫_0^2 x^2 dx = [x^3/3]_0^2 = 8/3 square units.

Use this result when the required region lies between a single curve and the x-axis over a fixed interval. The integrand represents the height of a vertical strip, and dx represents its small width. If the curve goes below the x-axis, ∫_a^b f(x) dx gives signed area, so actual area needs separate intervals or absolute value handling. Key rule: Area = ∫_a^b y dx = ∫_a^b f(x) dx when the region is bounded by y = f(x), x-axis, x = a, x = b, and f(x) ≥ 0 on [a, b]. If f(x) ≤ 0, actual area = -∫_a^b f(x) dx. If f(x) changes sign, split the interval at zeros of f(x). Find the area bounded by y = 4 - x^2 and the x-axis. Intersections with x-axis: 4 - x^2 = 0 gives x = -2, 2. Since 4 - x^2 ≥ 0 on [-2, 2], area = ∫_-2^2 (4 - x^2) dx. Using symmetry, area = 2∫_0^2 (4 - x^2) dx = 2[4x - x^3/3]_0^2 = 2(8 - 8/3) = 2(16/3) = 32/3 square units. Usually asked as a short or long answer question requiring limits from given boundaries, direct integration, and final area in square units. It may also appear with a curve and two vertical lines, where no intersection solving is needed. Using ∫_-2^2 x^2 - 4 dx for y = 4 - x^2 gives a negative value, because the integrand has been reversed relative to the shaded region.