Difficulty: Intermediate | Prerequisites: Definite integrals, basic antiderivatives, graphing functions, and the trig/IBP techniques from earlier in HW 05.
Areas between curves and the Fundamental Theorem of Calculus are the conceptual backbone of integral calculus. Computing areas between curves asks you to set up an integral that captures the vertical (or horizontal) distance between two functions across an interval, accounting for which function sits on top. The FTC ties everything together: Part 1 says differentiation undoes integration, and Part 2 says you can evaluate a definite integral using any antiderivative. Mastering these ideas means you understand why integration works, not just how to carry out the mechanics.
Area between two curves = ∫ (top - bottom) dx (or ∫ (right - left) dy for horizontal slicing). When a function crosses the x-axis, use absolute values or split the integral at the zeros. The FTC with the chain rule says (d/dx) ∫ₐ^{g(x)} f(t) dt = f(g(x)) g'(x).
Area between curves (vertical slicing)
Area = ∫ₐᵇ [f(x) - g(x)] dx, where f(x) >= g(x) on [a, b]. Think of it as: stacking thin vertical rectangles of height (top curve minus bottom curve) and summing their areas.
Area between curves (horizontal slicing)
Area = ∫_c^d [f(y) - g(y)] dy, where you integrate with respect to y. In simple terms, this is the same idea rotated 90 degrees, useful when curves are easier to describe as x = f(y).
Absolute area (signed vs unsigned)
When a function dips below the x-axis, ∫ f(x) dx gives a negative contribution. To find the actual geometric area, use ∫ |f(x)| dx, which means splitting the integral at zeros and negating where the function is negative.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x). In simple terms, differentiating an integral with a variable upper limit gives back the integrand.
FTC1 with the chain rule
If F(x) = ∫ₐ^{g(x)} f(t) dt, then F'(x) = f(g(x)) g'(x). The chain rule multiplies by the derivative of the upper limit.
Fundamental Theorem of Calculus, Part 2 (FTC2)
∫ₐᵇ f(x) dx = F(b) - F(a), where F is any antiderivative of f. This is the evaluation theorem that lets you compute definite integrals without Riemann sums.
When f(x) crosses the x-axis on the interval, the integral ∫ f(x) dx counts area below the axis as negative. To find the geometric area, take ∫ |f(x)| dx by splitting at the zeros.
Example (Q3): Area of the region bounded by f(x) = 5 sin³(x) on [0, 3π/2]. Since sin³(x) >= 0 on [0, π] and sin³(x) <= 0 on [π, 3π/2], split: Area = ∫₀^π 5 sin³(x) dx - ∫_π^{3π/2} 5 sin³(x) dx. Use the identity sin²(x) = 1 - cos²(x) and substitute u = cos(x). First integral = 20/3, second = -10/3. Area = 20/3 + 10/3 = 10.
Find where the curves intersect to determine the limits of integration. Then integrate (top - bottom).
Example (Q17): f(x) = 3x - x² + 5 and g(x) = x + 5. Set equal: 3x - x² + 5 = x + 5, so 2x - x² = 0, giving x = 0 and x = 2. Area = ∫₀² (2x - x²) dx = [x² - x³/3]₀² = 4/3.
When curves are given as x = f(y) and x = g(y), integrate with respect to y.
Example (Q18): f(y) = y² and g(y) = 2 - y. Intersection: y² = 2 - y, so y² + y - 2 = (y + 2)(y - 1) = 0, giving y = -2 and y = 1. Area = ∫{-2}^{1} (2 - y - y²) dy = [2y - y²/2 - y³/3]{-2}^{1} = 9/2.
Some problems ask you to express the area rather than compute it.
Example (Q19): Area between f(x) = eˣ, g(x) = e^{-2x}, and the line y = 3. The curves intersect y = 3 at x = ln 3 and x = -(ln 3)/2. The area is ∫_{-(ln 3)/2}^{0} (3 - e^{-2x}) dx + ∫₀^{ln 3} (3 - eˣ) dx.
The FTC says ∫ₐᵇ f'(x) dx = f(b) - f(a). If you know f is linear with slope m, then f'(x) = m.
Example (Q20): f(x) = mx + b, so f'(x) = m. ∫₀⁴ f'(x) dx = ∫₀⁴ m dx = 4m.
When the upper limit of integration is a function of x, apply the chain rule.
Example (Q21): F(x) = ∫₃^{√x} (8 sin t / t) dt. By FTC1 + chain rule: F'(x) = (8 sin(√x) / √x) (1/(2√x)) = 4 sin(√x) / x.
Example (Q22): ∫₀^{π/2} (cos x + 3 sin x) dx. Antiderivative: sin x - 3 cos x. Evaluate: (1 - 0) - (0 - 3) = 4.
\text{Area} = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx \quad \text{where } f(x) \geq g(x)\text{Area} = \int_a^b |f(x)|\,dx \quad \text{(geometric area when } f \text{ changes sign)}\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f\bigl(g(x)\bigr)\,g'(x)\int_a^b f'(x)\,dx = f(b) - f(a)Areas between curves appear in economics when calculating consumer and producer surplus (the area between a demand curve and a supply curve). The FTC is the theoretical foundation for every application of integration: whenever a rate of change is integrated to recover a total quantity (distance from velocity, total cost from marginal cost, population from growth rate), the FTC is doing the work.
Students often forget to take the absolute value when computing geometric area. If f(x) dips below the axis, ∫ f(x) dx alone gives the wrong (smaller) answer.
When finding the area between two curves, students sometimes integrate (bottom - top) and get a negative number. Always identify which function is larger on the interval.
In FTC1 with the chain rule, students forget to multiply by g'(x). The derivative of ∫ₐ^{g(x)} f(t) dt is f(g(x)) times g'(x), not just f(g(x)).
Setting up the wrong limits of integration. For area problems, the limits come from the intersection points of the curves, not from arbitrary values.
⚠️ Area problems that require splitting the integral (function crosses the axis or curves switch which is on top) appear regularly. Sketch the graph first to see where to split.
⚠️ FTC1 with the chain rule (Q21 type) is a classic exam problem. Memorise the formula: f(g(x)) g'(x).
⚠️ Problems asking you to express (not evaluate) the area as definite integrals (Q19 type) test whether you understand the setup. Getting the limits and the integrand right is the whole challenge.
⚠️ For area between curves, always find the intersection points algebraically. Do not assume them from a graph.
True or False: ∫₀^{2π} sin(x) dx gives the total area between sin(x) and the x-axis on [0, 2π]. (False: it gives 0, because the positive and negative parts cancel. You need ∫ |sin(x)| dx.)
Fill in the blank: (d/dx) ∫₃^{x²} f(t) dt = ______. (f(x²) 2x)
True or False: To find the area between y = x² and y = x, you first need to find where they intersect. (True: they intersect at x = 0 and x = 1.)
Fill in the blank: ∫₀⁴ f'(x) dx = ______. (f(4) - f(0))
True or False: When integrating with respect to y to find area, the formula uses (right curve - left curve). (True.)
Q: Find the area of the region bounded by f(x) = 5 sin³(x) on [0, 3π/2] and the x-axis.
A: Split at x = π where sin changes sign. ∫₀^π 5 sin³(x) dx = 20/3 (using u = cos x). ∫_π^{3π/2} 5 sin³(x) dx = -10/3. Area = 20/3 + 10/3 = 10.
Q: Find the area enclosed by f(x) = 3x - x² + 5 and g(x) = x + 5.
A: Intersections at x = 0 and x = 2 (from 2x - x² = 0). Area = ∫₀² (2x - x²) dx = [x² - x³/3]₀² = 4/3.
Q: Find the area enclosed by f(y) = y² and g(y) = 2 - y.
A: Intersections: y² + y - 2 = 0, so y = -2 and y = 1. Area = ∫_{-2}^{1} (2 - y - y²) dy = 9/2.
Q: Determine F'(x) when F(x) = ∫₃^{√x} (8 sin t / t) dt.
A: By FTC1 + chain rule: F'(x) = (8 sin(√x) / √x)(1/(2√x)) = 4 sin(√x) / x.
Q: Evaluate ∫₀^{π/2} (cos x + 3 sin x) dx.
A: Antiderivative: sin x - 3 cos x. Evaluate: [sin(π/2) - 3 cos(π/2)] - [sin 0 - 3 cos 0] = (1 - 0) - (0 - 3) = 4.
Areas between curves lay the groundwork for probability, where the area under a probability density function equals 1 and areas between bounds give probabilities. The FTC connects forward to the Mean Value Theorem for Integrals and to the computation of average values of functions. Both topics also feed into improper integrals (what happens when one limit goes to infinity?) and multivariable calculus, where areas generalise to double integrals.
area between curves, area under curve, absolute value integral, signed area, geometric area, Fundamental Theorem of Calculus, FTC Part 1, FTC Part 2, chain rule FTC, variable upper limit, intersection points, vertical slicing, horizontal slicing, consumer surplus, producer surplus, Stewart 5e, UT Austin integral calculus, HW 05 solutions