Area Between Two Curves

1. Write equations of both curves. 2. Solve them simultaneously to find intersection points. 3. Sketch both curves and shade the enclosed region. 4. Decide vertical or horizontal strips. 5. For dx, identify top minus bottom; for dy, identify right minus left. 6. Check whether the order stays the same over the interval. 7. Integrate, substitute limits, simplify and write square units.

Draw both curves on the same axes, mark their intersection points, shade only the enclosed region, and label the upper and lower curve on each interval. If using dy, label right and left boundaries instead.

For y = x and y = x^2, writing ∫_0^1 (x^2 - x) dx gives -1/6, which is signed area and cannot be the final area of a region.

The area between two curves y = f(x) and y = g(x) from x = a to x = b is ∫_a^b [upper curve - lower curve] dx, provided the same curve stays above on the whole interval.

The area between two curves y = f(x) and y = g(x) from x = a to x = b is ∫_a^b [upper curve - lower curve] dx, provided the same curve stays above on the whole interval. Key rule: Area = ∫_a^b [f(x) - g(x)] dx when f(x) ≥ g(x) on [a, b]. If using horizontal strips, Area = ∫_c^d [right curve - left curve] dy. Conditions: limits must match the bounded region, intersections must be found when not given, and the order must be changed or split if the curves cross. The area between y = x and y = x^2 from x = 0 to x = 1 is ∫_0^1 (x - x^2) dx = [x^2/2 - x^3/3]_0^1 = 1/2 - 1/3 = 1/6 square unit.

This concept applies when a closed region is formed by two curves or by curves along with given boundaries. The first task is to find intersection points, because they often become limits. After that, decide which curve is above the other on the interval. If the curves cross inside the interval, split the integral at the crossing point. Key rule: Area = ∫_a^b [f(x) - g(x)] dx when f(x) ≥ g(x) on [a, b]. If using horizontal strips, Area = ∫_c^d [right curve - left curve] dy. Conditions: limits must match the bounded region, intersections must be found when not given, and the order must be changed or split if the curves cross. Find the area enclosed between y = x and y = x^2. Intersections: x = x^2 gives x(x - 1) = 0, so x = 0 and x = 1. On (0, 1), x > x^2, so upper curve is y = x and lower curve is y = x^2. Area = ∫_0^1 (x - x^2) dx = [x^2/2 - x^3/3]_0^1 = 1/2 - 1/3 = 1/6 square unit. Often asked as a long answer problem where students must solve intersections, sketch the region, choose the correct subtraction order, integrate a polynomial or standard expression, and state the final area. For y = x and y = x^2, writing ∫_0^1 (x^2 - x) dx gives -1/6, which is signed area and cannot be the final area of a region.