What if the curves cross multiple times?

Split the integral into separate sub-intervals at each intersection point. Set up a distinct integral for each sub-interval, always subtracting the lower curve from the upper curve for that specific section, and sum the results.

How do I handle functions given in terms of y?

Integrate with respect to $y$. The area formula becomes $A = \int_{c}^{d} [f(y) - g(y)] dy$, where $f(y)$ is the rightmost curve and $g(y)$ is the leftmost curve.

What if the area is unbounded?

If the region extends infinitely in one or both directions, set up an improper integral with limits approaching infinity. If the integral converges to a finite value, that is the area; if it diverges, the area is infinite.

How to find the area between two curves

The area between two curves is found by integrating the difference between the upper function and the lower function over a specified interval. This method applies when both curves are continuous on the interval and is adapted by splitting the integral at intersection points if the curves cross.

The setup

Identify the two functions bounding the region. If the limits of integration are not given, find the points of intersection by setting the two equations equal to each other and solving for the variable of integration. Determine which curve is the upper bound and which is the lower bound on the interval by testing a point between the limits.

The steps

Determine the limits of integration, $a$ and $b$ . 2. Identify the upper function $f(x)$ and lower function $g(x)$ such that $f(x) \geq g(x)$ on $[a, b]$ . 3. Set up the definite integral: $A = \int_{a}^{b} [f(x) - g(x)] dx$ . 4. Find the antiderivative and evaluate it from $a$ to $b$ using the Fundamental Theorem of Calculus.

Checking the result

The computed area must be strictly non-negative. If the result is negative, the upper and lower functions were likely swapped, or an arithmetic error occurred during evaluation.

Common errors

A frequent error is failing to split the integral at intersection points where the upper and lower curves change roles. Another is integrating with respect to $x$ when the region is more easily defined by functions of $y$ , which leads to overly complex integrals.

Worked example

Find the area of the region bounded by $y = x^2$ and $y = 2x$ .

Set the equations equal to find intersection points: $x^2 = 2x \Rightarrow x^2 - 2x = 0 \Rightarrow x(x - 2) = 0$ . The limits are $x = 0$ and $x = 2$ . Test $x = 1$ : $2(1) = 2$ and $(1)^2 = 1$ , so $y = 2x$ is the upper curve. Set up the integral: $A = \int_{0}^{2} (2x - x^2) dx$ . Integrate: $[x^2 - \frac{1}{3}x^3]_0^2$ . Evaluate at bounds: $(2^2 - \frac{1}{3}(2)^3) - (0 - 0) = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$ . The area is $\frac{4}{3}$ .

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References: Stewart Calculus, Chapter 6: Applications of Integration · OpenStax Calculus Volume 1, Chapter 6: Applications of Integration · Khan Academy, AP Calculus AB: Area between curves