Find a region before integrating its height

Try

Sketch y=x and y=x² between zero and one. At x=1/2, the line has height 1/2 and the parabola height 1/4. A thin vertical strip between them has height 1/4 at that input. Its contribution to area is strip height times a small width, not the product of the two curve heights.

Predict

Predict where the curves meet by solving x=x². Then predict whether line-minus-parabola stays nonnegative between those intersections. If you extend the interval beyond 1, the order changes; one unsplit signed integral would no longer measure total area.

Experiment

The diagram uses equal horizontal and vertical scales. The vertical segment at x=1/2 represents the slice whose height is x−x².

Integrate the height: ∫₀¹(x−x²)dx=1/2−1/3=1/6 square units. The small area is plausible because the curves meet at both ends and stay close between them. A result larger than the full unit square would signal a setup or arithmetic error.

Understand

For a region with upper boundary u(x) and lower boundary v(x), vertical slices give area ∫[u(x)−v(x)]dx over the appropriate interval. The height must be nonnegative. Find intersections and other boundary changes before selecting the limits. Where the order changes, split the region or integrate the absolute difference.

For y=2x and y=x², intersections occur at x=0 and x=2. The area is ∫₀²(2x−x²)dx=4/3. A test point helps determine the order, but a factored difference x(2−x) proves nonnegativity throughout the interval. Checking only one point is not enough if additional intersections or discontinuities have been overlooked.

Horizontal slicing sometimes avoids complicated inverses. Between x=y² and x=2−y, solve y²=2−y to get y=−2 and y=1. The line lies to the right, so area is ∫₋₂¹(2−y−y²)dy=9/2. The differential dy tells you that horizontal width, not vertical height, belongs in the integrand.

Units provide another check. Boundary differences have length units, and strip thickness has length units, so their product gives area. If the coordinates instead describe time and rate, the same integral structure may represent a physical accumulation rather than a geometric land area.

Apply

For every practice region, sketch its intersections, label one representative strip and state why its height or width is nonnegative. Repair the signed-cancellation example by splitting at zero. Keep exact fractions until the final interpretation.

Continue

The next lesson applies the same sign distinction to motion: integrating velocity gives displacement, while integrating its magnitude gives distance traveled.