Finding Total Area with Curves Below the x-Axis

Evaluate the definite integral of f(x) = -3 from x = 1 to x = 4, then state the geometric area bounded by the curve and the x-axis.

The definite integral is -9. The geometric area is |-9| = 9 square units.

Find the geometric area bounded by f(x) = 2x - 6 and the x-axis on the interval [0, 3].

The line lies entirely below the axis on [0, 3]. Integrating gives [x^2 - 6x] from 0 to 3 = -9. Geometric area = 9 square units.

Find the x-intercepts of f(x) = x^2 - 1 on the interval [0, 2], and identify which subinterval lies below the x-axis.

The intercept is x = 1. The subinterval [0, 1] lies below the x-axis; [1, 2] lies above.

Calculate the total area bounded by f(x) = x^2 - 1 and the x-axis between x = 0 and x = 2.

Integral on [0, 1] is -2/3, giving area 2/3. Integral on [1, 2] is 4/3. Total area = 2/3 + 4/3 = 2 square units.

Find the total area bounded by f(x) = x^3 and the x-axis between x = -2 and x = 2.

The curve crosses at x = 0. Integral from -2 to 0 is -4 (area 4); integral from 0 to 2 is 4 (area 4). Total area = 4 + 4 = 8 square units.

Explain why evaluating the single integral of f(x) = x^3 from -2 to 2 gives 0 instead of the geometric area.

The region on [-2, 0] evaluates to -4 and the region on [0, 2] evaluates to +4. Because definite integrals calculate net signed area, the equal positive and negative regions cancel out.