# Area Between Curves

Mathematics I · Integral Calculus · https://tryals.app/learn/mathematics-i/area-between-curves

## Finding Enclosed Area

The **area between two curves** generalizes the area under a curve by integrating thin vertical strips across the region. If $f(x) \ge g(x)$ on $[a, b]$, the trapped area is:

$$A = \int_a^b \big(f(x) - g(x)\big)\, dx$$

The key rule is **top minus bottom**: the upper curve minus the lower one, ensuring the strip height is nonnegative. Unlike a single definite integral, which computes signed area, between-curves area is **always positive** because you subtract the lower function from the upper, making the $x$-axis completely irrelevant.

| Step | Action | Description |
| --- | --- | --- |
| **1** | **Intersections** | Solve $f(x) = g(x)$ for limits $a$ and $b$. |
| **2** | **Top/Bottom** | Identify which curve is higher over the interval. |
| **3** | **Integrate** | Evaluate $\int (\text{top} - \text{bottom})\, dx$. |

## Pitfalls and Alternatives

A major **common pitfall** is integrating without checking curve order or ignoring crossings. If $f$ and $g$ **cross** inside $[a, b]$, you must **split the integral** at the crossing point and compute top-minus-bottom on each sub-interval separately. Subtracting in the wrong order yields a negative area.

| Variable | Orientation | Strip Formula |
| --- | --- | --- |
| **x-axis** | Vertical strips | $\int \big(\text{top} - \text{bottom}\big)\, dx$ |
| **y-axis** | Horizontal strips | $\int \big(\text{right} - \text{left}\big)\, dy$ |

When curves are functions of $y$, integrate horizontally using **right minus left** strips.

## Practice questions

4 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Select every statement about the area between two curves that is true.

A. The formula fails whenever either curve lies below the x-axis.
B. When the curves cross, the interval must be split at the crossing.
C. Where the curves meet marks the ends of the region.
D. The integrand is the vertical gap: top curve minus bottom curve.
E. Where a region is easier described by its left and right edges, integrate in y instead.

**Answer:** A. The formula fails whenever either curve lies below the x-axis.; B. When the curves cross, the interval must be split at the crossing.; D. The integrand is the vertical gap: top curve minus bottom curve.; E. Where a region is easier described by its left and right edges, integrate in y instead.

**Why:** All are true except the second: subtracting one curve from the other measures the gap between them, and a gap is positive wherever the top curve really is on top, regardless of where the axis lies.

Page: https://tryals.app/practice/mathematics-i/area-between-curves/select-every-statement-about-the-area-between-two-curves-that-is-true

### 2. Order the steps for finding the area of the region enclosed between two curves.

**Answer:**

1. Solve f(x) = g(x) to find where the curves intersect
2. Take the intersection points as the limits of integration
3. Decide which curve is the upper one on each interval
4. Integrate top curve minus bottom curve across each interval
5. Add the pieces to get the total enclosed area

**Why:** Intersections give the limits; the ordering of the curves gives the integrand. Skipping the third step is what produces negative 'areas', and skipping the possibility of a swap is what loses half of a region.

Page: https://tryals.app/practice/mathematics-i/area-between-curves/order-the-steps-for-finding-the-area-of-the-region-enclosed-between

### 3. To find the area between two curves that cross, you must split the integral at the crossing point. What does a single integral across the whole interval actually compute?

A. The net signed area, so the parts partly cancel
B. Zero, because the two regions are always equal
C. The correct area, since the formula handles crossings automatically
D. The area of the larger region only

**Answer:** A. The net signed area, so the parts partly cancel

**Why:** Past the crossing the integrand changes sign, so the second region subtracts from the first. The integral is doing exactly what it always does, signed area, and the splitting is how you tell it you wanted geometric area instead, the same distinction as displacement against distance.

Page: https://tryals.app/practice/mathematics-i/area-between-curves/to-find-the-area-between-two-curves-that-cross-you-must-split-the

### 4. The area between two curves is ∫(top − bottom) dx. Why does that formula work even when both curves dip below the x-axis, where each individual integral would be negative?

A. Because curves below the axis must first be reflected upward before the formula applies.
B. Because the two negative integrals multiply to give a positive result.
C. Because the subtraction measures the vertical gap between the curves.
D. Because areas are defined to be positive, so the negative signs are discarded at the end.

**Answer:** C. Because the subtraction measures the vertical gap between the curves.

**Why:** Test it with f = −1 above g = −3: the individual integrals are −1 and −3, and their difference is 2, the correct area. The integrand is the height of a strip, and where each curve sits relative to the axis never enters. What does need care is curves that *cross*, since the integrand changes sign, which is why finding the intersections comes first.

Page: https://tryals.app/practice/mathematics-i/area-between-curves/the-area-between-two-curves-is-top-bottom-dx-why-does-that