The Definite Integral for MATH 122
Exam Relevance for MATH 122
Tested through FTC applications. Understanding as accumulated change is important.
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Introduction
Now that you know how to find anti-derivatives, it's time to use them to calculate something concrete: area under a curve.
The definite integral connects anti-derivatives to geometry. While an indefinite integral gives you a family of functions (with that $+C$), a definite integral gives you a single number — typically representing an area.
Key Insight: The definite integral $\int_a^b f(x)\,dx$ answers the question: "What is the signed area between the curve $y = f(x)$ and the x-axis, from $x = a$ to $x = b$?"
Notation and Terminology
The definite integral is written as:
$$\int_a^b f(x)\,dx$$
where:
- $a$ is the lower limit of integration
- $b$ is the upper limit of integration
- $f(x)$ is the integrand (the function being integrated)
- $dx$ indicates we're integrating with respect to $x$
The numbers $a$ and $b$ are also called the bounds or limits of integration.
Evaluating Definite Integrals
To evaluate a definite integral:
- Find an anti-derivative $F(x)$ of the integrand
- Plug in the upper limit: $F(b)$
- Plug in the lower limit: $F(a)$
- Subtract: $F(b) - F(a)$
Notation shortcut: We write $F(x)\Big|_a^b$ or $\left[F(x)\right]_a^b$ to mean "$F(x)$ evaluated from $a$ to $b$."
Evaluate $\int_1^4 x^2\,dx$.
Step 1: Find an anti-derivative
Using the power rule for integration:
$$F(x) = \frac{x^3}{3}$$
(We don't need the $+C$ for definite integrals — it cancels out!)
Step 2: Evaluate at the bounds
$$\int_1^4 x^2\,dx = \frac{x^3}{3}\Big|_1^4 = \frac{4^3}{3} - \frac{1^3}{3}$$
$$= \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21$$
Answer: $\int_1^4 x^2\,dx = 21$
Signed Area: Above vs. Below the x-axis
Critical concept: The definite integral computes signed area:
- Area above the x-axis counts as positive
- Area below the x-axis counts as negative
This is different from "total area" which counts everything as positive.
Evaluate $\int_0^{\pi} \sin(x)\,dx$ and $\int_{\pi}^{2\pi} \sin(x)\,dx$.
Part A: $\int_0^{\pi} \sin(x)\,dx$
Step 1: Find the anti-derivative
The anti-derivative of $\sin(x)$ is $-\cos(x)$.
Step 2: Evaluate at the bounds
$$\int_0^{\pi} \sin(x)\,dx = -\cos(x)\Big|_0^{\pi}$$
$$= -\cos(\pi) - (-\cos(0))$$
$$= -(-1) - (-1)$$
$$= 1 + 1 = 2$$
This is positive because $\sin(x)$ is above the x-axis on $[0, \pi]$.
Part B: $\int_{\pi}^{2\pi} \sin(x)\,dx$
Step 1: Find the anti-derivative
The anti-derivative of $\sin(x)$ is $-\cos(x)$.
Step 2: Evaluate at the bounds
$$\int_{\pi}^{2\pi} \sin(x)\,dx = -\cos(x)\Big|_{\pi}^{2\pi}$$
$$= -\cos(2\pi) - (-\cos(\pi))$$
$$= -1 - (-(-1))$$
$$= -1 - 1 = -2$$
This is negative because $\sin(x)$ is below the x-axis on $[\pi, 2\pi]$.
Key observation: Notice that $\int_0^{2\pi} \sin(x)\,dx = 2 + (-2) = 0$. The positive and negative areas cancel out!
Properties of Definite Integrals
These properties help simplify calculations:
1. Constant Multiple: $$\int_a^b cf(x)\,dx = c\int_a^b f(x)\,dx$$
2. Sum/Difference: $$\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$$
3. Splitting the Interval: $$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$$
4. Reversing Limits: $$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$$
5. Same Limits: $$\int_a^a f(x)\,dx = 0$$
Given that $\int_0^3 f(x)\,dx = 7$ and $\int_0^3 g(x)\,dx = -2$, find $\int_0^3 [2f(x) - 3g(x)]\,dx$.
Step 1: Apply the sum/difference property
$$\int_0^3 [2f(x) - 3g(x)]\,dx = \int_0^3 2f(x)\,dx - \int_0^3 3g(x)\,dx$$
Step 2: Apply the constant multiple property
$$= 2\int_0^3 f(x)\,dx - 3\int_0^3 g(x)\,dx$$
Step 3: Substitute the given values
$$= 2(7) - 3(-2) = 14 + 6 = 20$$
Answer: $\int_0^3 [2f(x) - 3g(x)]\,dx = 20$
Area Between a Curve and the x-axis
To find the actual area (always positive), you need to:
- Identify where the function crosses the x-axis
- Integrate each piece separately
- Take the absolute value of negative pieces
- Add everything up
Find the total area between $f(x) = x^2 - 4$ and the x-axis from $x = 0$ to $x = 3$.
Step 1: Find where $f(x) = 0$
$$x^2 - 4 = 0$$
$$x^2 = 4$$
$$x = \pm 2$$
So $f(x) = 0$ at $x = 2$ (within our interval $[0, 3]$).
Step 2: Determine which parts are above/below
- For $0 \le x < 2$: Test $f(1) = 1 - 4 = -3 < 0$ → below x-axis
- For $2 < x \le 3$: Test $f(3) = 9 - 4 = 5 > 0$ → above x-axis
Step 3: Integrate each piece
Below the axis (take absolute value):
$$\left|\int_0^2 (x^2 - 4)\,dx\right| = \left|\frac{x^3}{3} - 4x\Big|_0^2\right|$$
$$= \left|\left(\frac{8}{3} - 8\right) - (0)\right| = \left|-\frac{16}{3}\right| = \frac{16}{3}$$
Above the axis:
$$\int_2^3 (x^2 - 4)\,dx = \frac{x^3}{3} - 4x\Big|_2^3$$
$$= \left(\frac{27}{3} - 12\right) - \left(\frac{8}{3} - 8\right)$$
$$= (9 - 12) - \left(\frac{8}{3} - 8\right)$$
$$= -3 - \frac{8}{3} + 8 = 5 - \frac{8}{3} = \frac{7}{3}$$
Step 4: Add the areas
$$\text{Total area} = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}$$
Answer: The total area is $\frac{23}{3} \approx 7.67$ square units.
Conceptual Check: True or False?
Test your understanding with these true/false questions.
True or False: $\int_a^b f(x)\,dx$ is always positive.
False. The definite integral gives signed area. If the function is below the x-axis, the integral will be negative.
True or False: If $f(x) > 0$ for all $x$ in $[a, b]$, then $\int_a^b f(x)\,dx > 0$.
True. When a function is entirely above the x-axis on an interval, the area under it (and hence the integral) is positive.
True or False: $\int_0^2 x\,dx = \int_2^0 x\,dx$.
False. Reversing the limits changes the sign. We have $\int_0^2 x\,dx = -\int_2^0 x\,dx$.
Let's verify:
$$\int_0^2 x\,dx = \frac{x^2}{2}\Big|_0^2 = 2 - 0 = 2$$
$$\int_2^0 x\,dx = \frac{x^2}{2}\Big|_2^0 = 0 - 2 = -2$$
Indeed, they are negatives of each other.
True or False: If $\int_a^b f(x)\,dx = 0$, then $f(x) = 0$ for all $x$ in $[a, b]$.
False. The integral can be zero if positive and negative areas cancel out. For example, $\int_0^{2\pi} \sin(x)\,dx = 0$, but $\sin(x)$ is not always zero — it's positive on $[0, \pi]$ and negative on $[\pi, 2\pi]$, and these areas cancel.
More Examples
Evaluate $\int_0^2 (3x^2 + 2x - 1)\,dx$.
Step 1: Find the anti-derivative
$$F(x) = x^3 + x^2 - x$$
Step 2: Evaluate at bounds
$$\int_0^2 (3x^2 + 2x - 1)\,dx = x^3 + x^2 - x\Big|_0^2$$
$$= (8 + 4 - 2) - (0 + 0 - 0) = 10 - 0 = 10$$
Answer: $\int_0^2 (3x^2 + 2x - 1)\,dx = 10$
Evaluate $\int_0^1 e^x\,dx$.
Step 1: Find the anti-derivative
The anti-derivative of $e^x$ is $e^x$.
Step 2: Evaluate at bounds
$$\int_0^1 e^x\,dx = e^x\Big|_0^1 = e^1 - e^0 = e - 1$$
Answer: $\int_0^1 e^x\,dx = e - 1 \approx 1.718$
Evaluate $\int_{-2}^{2} |x|\,dx$.
Step 1: Understand the function
$|x| = \begin{cases} -x & \text{if } x < 0 \\ x & \text{if } x \ge 0 \end{cases}$
Step 2: Split at the point where the formula changes
$$\int_{-2}^{2} |x|\,dx = \int_{-2}^{0} (-x)\,dx + \int_{0}^{2} x\,dx$$
Step 3: Evaluate each piece
$$\int_{-2}^{0} (-x)\,dx = -\frac{x^2}{2}\Big|_{-2}^{0} = 0 - \left(-\frac{4}{2}\right) = 2$$
$$\int_{0}^{2} x\,dx = \frac{x^2}{2}\Big|_{0}^{2} = 2 - 0 = 2$$
Step 4: Add
$$\text{Total} = 2 + 2 = 4$$
Answer: $\int_{-2}^{2} |x|\,dx = 4$
Geometric check: This represents two triangles, each with base 2 and height 2, so area $= 2 \times \frac{1}{2}(2)(2) = 4$. ✓
Definite Integral Notation
The definite integral equals the anti-derivative evaluated at the upper limit minus the anti-derivative evaluated at the lower limit.
Variables:
- $a$:
- lower limit of integration
- $b$:
- upper limit of integration
- $f(x)$:
- the integrand (function being integrated)
- $F(x)$:
- any anti-derivative of f(x)
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