Area Between Curves (Functions of x and of y)

Area between curves. Vertical strips: $\int_a^b (\text{top} - \text{bottom})\,dx$, with $a$ and $b$ the $x$-values of the ends. Horizontal strips: $\int_c^d (\text{right} - \text{left})\,dy$, with $c$ and $d$ the $y$-values of the ends. The limits are where the curves meet (or the lines that close the region); on the calculator part of the AP exam they are found with the calculator and stored, and the integral is written out before its value.

The usual slips: bottom minus top (a negative "area"); left minus right in a $dy$ integral; $x$-values used as the limits of a $dy$ integral; a meeting point missed, so the region is cut short; and one integral used where the top or bottom curve changes partway (then split the region, or turn to $dy$).

1. A parabola and a line. Find the area of the region enclosed by $y = x^{2}$ and $y = 2x + 3$.

Ends. Set the curves equal: $x^2 = 2x + 3$, so $x^2 - 2x - 3 = 0$ and $x = -1$ or $x = 3$.

Top and bottom. At $x = 0$, between the ends, the line is at 3 and the parabola at 0: the line is the top.

The area is $\frac{32}{3}$ square units. Check: the result is positive, as an area must be; the reverse order would give $-\frac{32}{3}$.

2. Meeting points from the calculator. Let $R$ be the region in the first quadrant enclosed by $y = 3\sin x$ and $y = \frac{x^2}{2}$. Find the area of $R$.

Ends. The curves meet at $x = 0$, and the equation $3\sin x = \frac{x^2}{2}$ has no algebraic solution, so the calculator finds the other meeting point: $x = {2.201}$. Store it with all its digits; do not retype the rounded value.

Top and bottom. At $x = 1$, $3\sin 1 - \frac{1}{2} = {2.024} > 0$: the sine curve is the top.

The area of $R$ is about $2.991$. On the AP exam the integral with its limits is written down first, then the calculator's value to three decimal places.

3. Curves given as functions of $y$. Find the area of the region enclosed by $x = 4 - y^{2}$ and $x = y - 2$.

Both curves give $x$ in terms of $y$, and a vertical strip would run from the parabola's lower half to its upper half on one part and from the line on another. Horizontal strips run from the line to the parabola all the way, so use $dy$.

Ends. $4 - y^{2} = y - 2$ gives $y^{2} + y - 6 = 0$, so $y = -3$ or $y = 2$. These $y$-values are the limits.

Right and left. At $y = 0$ the parabola is at $x = 4$ and the line at $x = -2$: the parabola is on the right.

The area is $\frac{125}{6}$ square units. Using the $x$-values of the meeting points, $({-5}, {-3})$ and $({0}, {2})$, as the limits of this $dy$ integral would be wrong: in $dy$ the limits are heights.

4. A curve, a line and an axis: choosing $dx$ or $dy$. Find the area of the region bounded by $y = \sqrt{x}$, $y = x - 2$ and the $x$-axis.

With vertical strips the bottom changes at $x = 2$: the $x$-axis on $[0, 2]$, the line on $[2, 4]$. That needs two integrals. With horizontal strips every strip runs from the curve to the line. Solve each boundary for $x$: $y = \sqrt{x}$ is $x = y^{2}$ (left), $y = x - 2$ is $x = y + 2$ (right). The region runs from the $x$-axis, $y = 0$, up to where they meet: $y^{2} = y + 2$ gives $y = 2$ (the root $y = -1$ is below the axis, off this region).

Check with $dx$: $\int_0^2 \sqrt{x}\,dx + \int_2^4 \big(\sqrt{x} - (x - 2)\big)\,dx = {\frac{4\sqrt{2}}{3}} + {\frac{10 - 4\sqrt{2}}{3}} = {\frac{10}{3}}$, the same. A single integral $\int_0^4 \big(\sqrt{x} - (x - 2)\big)\,dx$ would wrongly add the triangle between the line and the $x$-axis on $[0, 2]$, which lies below the axis and is not part of the region.

5. A line that cuts a region in half (calculator). The region $R$ is bounded by $y = 4 - x^{2}$ and the $x$-axis. Find $k$ so that the horizontal line $y = k$ divides $R$ into two regions of equal area.

The whole area. $\int_{-2}^{2} (4 - x^2)\,dx = {\frac{32}{3}}$, so each part must have area $\frac{16}{3}$.

The part above $y = k$, in $dy$. Solving $y = 4 - x^{2}$ for $x$ gives $x = \pm\sqrt{4 - y}$, so a horizontal strip at height $y$ runs from $-\sqrt{4 - y}$ to $\sqrt{4 - y}$, length $2\sqrt{4 - y}$. Strips in $dy$ suit this problem, because the line $y = k$ is then simply a limit:

This equation in $k$ is solved on the calculator: $k = {1.480}$. (It is not $k = 2$: the region is wider at the bottom, so the line sits below the middle height.)