# The Definite Integral

Mathematics I · Integral Calculus · https://tryals.app/learn/mathematics-i/the-definite-integral

## Accumulating Under a Curve

The **definite integral** $\int_a^b f(x)\, dx$ measures the **accumulated (signed) area** between $y = f(x)$ and the $x$-axis from $a$ to $b$. While derivatives measure instantaneous rates, definite integrals measure **totals**, like total distance from velocity.

It is defined as the limit of **Riemann sums**, partitioning $[a,b]$ into $n$ subintervals of width $\Delta x$:

$$\int_a^b f(x)\, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i)\,\Delta x$$

Here, $\Delta x$ is subinterval width, $f(x_i)$ is sample height, and the integral sign is an elongated "S" for sum.

## Signed Area and Properties

| Property / Feature | Description | Result |
| --- | --- | --- |
| **Signed Area** | Area below the axis counts as **negative**. | Can be zero or negative. |
| **Integral Type** | Unlike indefinite integrals, this is a **number**. | **No $+C$** constant of integration. |
| **Reversing Limits** | $\int_b^a f(x)\, dx$ | Equals $-\int_a^b f(x)\, dx$. |
| **Zero Width** | $\int_a^a f(x)\, dx$ | Always equals $0$. |

> **Common Pitfall:** Treating definite integrals as purely geometric positive area. Parts below the axis subtract from the total.

## Practice questions

4 of this lesson's 10 practice questions, with answers. The full set is in the app.

### 1. The definite integral of a velocity that goes negative gives **displacement**, not distance travelled. Why does the integral come out smaller than the odometer?

A. Because displacement and distance are the same and one calculation must be wrong
B. Because the integral cannot handle negative values and truncates them
C. Because velocity should have been squared before integrating
D. Because area below the axis counts as negative and cancels

**Answer:** D. Because area below the axis counts as negative and cancels

**Why:** The signed area cancels: out and back gives zero displacement and twenty kilometres on the odometer. Integrating |v| recovers the distance, which is why the sign convention is not a nuisance but the thing that makes the integral answer 'where did it end up' rather than 'how far did it go'.

Page: https://tryals.app/practice/mathematics-i/the-definite-integral/the-definite-integral-of-a-velocity-that-goes-negative-gives

### 2. Select every statement about definite integrals that is true.

A. The integral of f from a to a is 0
B. A definite integral needs a constant of integration
C. The integral from a to c equals the integral from a to b plus the integral from b to c
D. The integral of f from b to a equals the integral from a to b
E. Area lying below the axis contributes negatively

**Answer:** A. The integral of f from a to a is 0; C. The integral from a to c equals the integral from a to b plus the integral from b to c; E. Area lying below the axis contributes negatively

**Why:** A strip of zero width accumulates nothing; adjacent intervals add; below-axis area subtracts. Reversing the limits _flips the sign_ rather than preserving it, and a definite integral is a number, so no + C appears.

Page: https://tryals.app/practice/mathematics-i/the-definite-integral/select-every-statement-about-definite-integrals-that-is-true

### 3. Order the steps that build the definite integral out of rectangles.

**Answer:**

1. Partition the interval from a to b into n subintervals of width delta x
2. Choose a sample point x_i in each subinterval and read off the height f(x_i)
3. Form the rectangle areas f(x_i) times delta x and add them up
4. Let n tend to infinity so the strips become arbitrarily thin
5. Identify the limit of the sums as the definite integral

**Why:** Partition, sample, sum, refine, take the limit. Each finite stage is only an approximation; the integral is what the approximations converge to, which is why the definition needs a limit rather than any single sum.

Page: https://tryals.app/practice/mathematics-i/the-definite-integral/order-the-steps-that-build-the-definite-integral-out-of-rectangles

### 4. A Riemann sum with more rectangles is always a better approximation than one with fewer.

**Answer:** False

**Why:** False. The *rule* matters too: a midpoint or trapezoid sum with four intervals routinely beats a left-endpoint sum with forty, because the endpoint rules have first-order error while the midpoint rule's cancels to second order. More rectangles help within a rule, not across rules.

Page: https://tryals.app/practice/mathematics-i/the-definite-integral/a-riemann-sum-with-more-rectangles-is-always-a-better-approximation