Back to the on-screen lesson ·
A vertical strip of length $f - g$, swept between the crossings, with the region split wherever the two curves swap places.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find the crossings of two curves and use them as the limits, test a point between consecutive crossings to decide which curve is on top rather than guessing, split the region wherever the ordering changes so that a total area is not reported as a net one, choose between vertical and horizontal strips by which direction meets each boundary once, and say why the position of the $x$-axis plays no part in the formula.
In Calculus I the definite integral $\int_a^b f$ was constructed as a limit of sums of rectangles, and where $f$ is positive it is the area under the graph. Where $f$ is negative, the rectangles have negative height and the integral counts that region negatively.
That sign convention is exactly right for accumulation — a rate that goes negative should undo what came before — and exactly wrong for area, which is never negative. Every difficulty in this lesson comes from that one mismatch.
A strip is a thin slice of the region, vertical (thickness $dx$) or horizontal (thickness $dy$). Its area is its length times its thickness, and the integral adds the strips up.
The gap is the strip's length: upper minus lower for a vertical strip, right minus left for a horizontal one.
The net signed area is what a single integral of $f - g$ gives, with regions where $g$ is above counting negatively. The total area counts every region positively, and it is what the word area means unless something says otherwise.
The formula. If $f \ge g$ on $[a, b]$, the area between the curves is $$A = \int_a^b \left(f(x) - g(x)\right)dx.$$ The integrand is the height of a vertical strip and $dx$ is its width. The hypothesis $f \ge g$ throughout is what makes this an area rather than a signed quantity.
The limits are not given. In almost every problem $a$ and $b$ are the crossings, found by solving $f(x) = g(x)$. Finding them is the first substantial step and the one most often rushed.
Which curve is on top is decided by testing a single point between consecutive crossings, not by which is written first or which looks bigger. Between two consecutive crossings the ordering cannot change — that is what a crossing is — so one test point settles the whole subinterval.
If the curves cross inside the interval, split there and integrate $|f - g|$ piece by piece, taking upper minus lower afresh on each. Otherwise you compute the net signed area, and the two can differ by everything: they differ by a factor of infinity when the net is zero.
Slicing the other way. Horizontal strips integrate $\left(x_{\text{right}}(y) - x_{\text{left}}(y)\right)dy$. Choose the direction in which each strip meets each boundary exactly once; when one direction needs the region split and the other does not, the other one is right.
No curve is needed above the axis. Nothing in the formula refers to the $x$-axis, so the region may sit anywhere: only the gap matters. Every application of the integral is the same move: write what one thin slice contributes, then add the slices up. Get the slice right and the integral writes itself; reach for a remembered formula instead and the first unfamiliar region defeats you.
Another way: picture
Two curves and a vertical line segment drawn between them, with the segment's foot on the lower curve and its head on the upper. That segment has length $f(x) - g(x)$ whatever height the pair sits at — shift both curves up by a hundred and the segment is unchanged. The area is the segment swept from one crossing to the other, which is why the axis plays no part.
Another way: steps
Consider $y = x^3$ and $y = x$ on $[-1, 1]$. They cross at $-1$, $0$ and $1$. On $(0,1)$ the line is above; on $(-1,0)$ the cubic is above.
The single integral $\displaystyle\int_{-1}^{1}(x - x^3)dx$ is zero, because the integrand is odd. As a statement about accumulation that is correct and meaningful: the two regions are congruent and count oppositely. As a statement about area it is absurd — there are plainly two regions of positive area there.
The total is $\displaystyle\int_{-1}^{0}(x^3 - x)dx + \int_{0}^{1}(x - x^3)dx = \tfrac14 + \tfrac14 = \tfrac12$.
So before writing anything down, decide which question is being asked. "The area enclosed" means total. "The net change" or "the displacement" means net. A problem that says neither usually wants total, and the giveaway that you have answered the wrong one is an answer of zero.
Using the given interval instead of the crossings. If a problem says "the region enclosed between", the interval is whatever the curves themselves enclose, and you have to find it.
Dividing by $x$ when solving $f = g$. $ax = x^2$ has solutions $0$ and $a$. Dividing by $x$ silently discards the first, which is usually one of the limits.
Assuming the ordering from the picture at one end. Curves that cross three times need three tests, not one.
Subtracting in a fixed order. Upper minus lower is decided per subinterval. Writing $f - g$ throughout and taking the absolute value at the end is not the same thing: $|\int| \ne \int|\cdot|$.
Slicing vertically out of habit. A sideways parabola has two branches; vertical strips need both and horizontal strips need neither.
Forgetting the region can sit below the axis. It changes nothing. The gap is a difference, and differences do not care about the axis.
The formula $\int (f - g)$ is usually explained as "the area under $f$ minus the area under $g$", and for a region sitting above the axis with both curves positive that description happens to give the right answer. It is the wrong picture, and it fails the moment the region moves.
If both curves are negative, there is no "area under" either of them in any ordinary sense — the integrals are negative — yet the difference is still exactly right. If one curve is above the axis and the other below, subtracting a negative integral adds, which looks like a trick and is not one.
The correct picture is a single vertical segment of length $f(x) - g(x)$, swept from one crossing to the other. The length of a segment depends only on the difference of its endpoints, so shifting both curves by any amount changes nothing — and the axis, which does not appear in the formula, plays no part in the geometry either.
Holding that picture also makes the horizontal case obvious rather than a second formula to memorise: turn your head, and the segment is horizontal with length $x_{\text{right}} - x_{\text{left}}$.
$y = x^2 - 2$ and $y = x$: solve $x^2 - 2 = x$, so $x^2 - x - 2 = (x-2)(x+1) = 0$ and the crossings are $-1$ and $2$.
The limits come from the curves, not from the question.
Test $x = 0$: the line is at $0$, the parabola at $-2$. The line is above, throughout $(-1, 2)$, since there is no crossing inside.
One test point per subinterval.
$\displaystyle\int_{-1}^{2}\left(x - x^2 + 2\right)dx = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3} + 2x\right]_{-1}^{2} = \dfrac{10}{3} - \left(-\dfrac{7}{6}\right) = \dfrac{9}{2}$.
Positive, as an area must be.
The region bounded by $y = x - 1$ and $y^2 = 2x + 6$. Vertical strips would need the parabola's two branches separately, and the region would split at $x = -3$.
Count the formulas each strip needs.
Horizontally: solve for $x$ in each, $x = y + 1$ and $x = \tfrac{y^2 - 6}{2}$, and they meet where $y = -2$ and $y = 4$. Each horizontal strip meets each boundary once.
One formula per end, for every $y$.
$\displaystyle\int_{-2}^{4}\left(y + 1 - \dfrac{y^2 - 6}{2}\right)dy = 18$ — one integral instead of two, and the same region.
Right minus left, this time.
Crossings: $x^2 = 2 - x^2$ gives $x^2 = 1$, so $x = \pm 1$.
Two crossings, so one subinterval.
Test $x = 0$: the second curve is at $2$ and the first at $0$, so $2 - x^2$ is on top throughout $(-1, 1)$.
One test point settles it.
$\displaystyle\int_{-1}^{1}\left(2 - 2x^{2}\right)dx = \left[2x - \dfrac{2x^3}{3}\right]_{-1}^{1} = \dfrac{8}{3}$.
Find the area enclosed between $y = 7x$ and $y = x^{2}$.
Answer:
Put the steps of finding the area enclosed between $y = 7x$ and $y = x^{2}$ into the order you do them.
Number the steps in order (write the number in the box):
Plot the two points where $y = 2x$ meets $y = x^{2}$.
Plot your answer on the grid:
On which interval is $y = 5x$ above $y = x^{2}$? Give the interval.
This task has no paper form; do it on a device.
A region is bounded on the left by $x = y^{2}$ and on the right by $x = 3$. Which way should you slice it?
Find the total area between $y = x^{3}$ and $y = 9x$ for $-3 \le x \le 3$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Find the area enclosed between $y = 4x$ and $y = x^{2}$.
Answer:
You can set up an area between curves without being given the limits, and say whether a question wants the net or the total. Say in your own words why the axis does not appear in the formula. Next: the same strip, spun about a line, becoming a volume.
10. Your turn: the area enclosed between $y = x^{2}$ and $y = 2 - x^{2}$, step 3
Positive, and symmetric as the picture demands.