Areas Between Two Curves
Find where two curves meet, then integrate the difference of the two functions between those limits to get the area enclosed.
What a learner can do afterwards
- Find and interpret intersections of a line and a parabola
- Evaluate geometric area using definite integrals, including separate signed pieces
- Choose top minus bottom for vertical strips or right minus left for horizontal strips
- Use supplied interval boundaries when an area is not enclosed solely by the curves
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Find intersections by setting the two rules equal. For y = 3x and y = x², x² - 3x = x(x - 3) = 0 gives x = 0 or 3. Substitute into the line to get (0, 0) and (3, 9). These are full points, not just integration limits.
For y = x + 12 and y = x², equating gives x² - x - 12 = (x - 4)(x + 3) = 0. The roots are 4 and -3. Compare their signed values to identify the smaller crossing x-coordinate.
This uses definite integration. On a <= x <= b where top and bottom keep their order, choose F with F'(x) = top - bottom and evaluate F(b) - F(a). If only the two curves enclose the region, their intersection x-coordinates give the outer limits. If vertical boundaries are supplied, use those instead. Check for crossings inside the interval and split where the order changes.
A vertical strip has height top minus bottom. For y = 2x and y = x², the enclosed limits are 0 and 2. Integrate 2x - x²: F(x) = x² - x³/3. The area is F(2) - F(0) = 4 - 8/3 = 4/3.
For y = 4x and y = x², the crossings are 0 and 4 and the line is higher between them. An antiderivative of 4x - x² is 2x² - x³/3. Evaluate at both limits: (32 - 64/3) - 0 = 32/3.
Choose the upper function by a sketch or a test value between consecutive crossings. On 0 < x < 1, x⁵ < x, so the height between y = x and y = x⁵ is x - x⁵. A test point must lie inside the interval, not at a crossing.
When two curves cross inside the interval, the upper and lower functions exchange places. Split there, subtract lower from upper on each piece, and add the positive areas. A fixed subtraction order would instead allow signed parts to cancel.
For y = x + 6 and y = x², the enclosed limits are -2 and 3. The line is higher. Integrate x + 6 - x² using F(x) = x²/2 + 6x - x³/3. F(3) = 27/2 and F(-2) = -22/3, so the area is 27/2 - (-22/3) = 125/6.
A supplied interval can bound an area even when its endpoints are not intersections. For y = 16 and y = x² restricted to 1 <= x <= 3, the limits are 1 and 3. The curves' intersections at -4 and 4 lie outside this restricted region.
A line tangent to a parabola has one distinct intersection. Equating y = x² with y = 8x - 16 gives (x - 4)² = 0. This contact does not by itself enclose a positive-area region.
A definite integral is signed. For y = -3 and the x-axis on [0, 2], integrating -3 gives -6, while the geometric area is 6. For area, subtract the lower function from the upper rather than accepting a negative answer.
Split wherever f - g changes sign. Within each resulting piece, its sign must stay constant before you turn a signed integral into geometric area by taking its magnitude. If two such pieces have signed integrals 2 and -7, their areas are 2 and 7, giving total area 9. If either piece still contains a sign change, subdivide it before adding magnitudes.
Horizontal strips use right minus left and integrate with respect to y. Between x = y² and x = 9 for 0 <= y <= 3, the strip width is 9 - y². The area setup is the integral of 9 - y² from y = 0 to 3.
Meeting is not always crossing. At a tangential contact, top and bottom may stay in the same order on both sides. For y = (x - 2)² and y = 0, the square stays nonnegative. Splitting at 2 is allowed but is not needed to prevent cancellation.
Use supplied boundaries or enclosing intersections for the limits. Integrate top minus bottom, splitting at every change of order, then add the nonnegative areas.
Lesson and assessment explanations written for Marble.
Additional questions and readings written for Marble. Videos credit Khan Academy.
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