The anti-derivative of a function $f(x)$ is a function $F(x)$ such that $F’(x) = f(x)$. The process of finding the anti-derivative is called anti-differentiation or integration.
The indefinite integral of $f(x)$ with respect to $x$ is denoted by:
$$\int f(x) \, dx = F(x) + c$$
where:
* $\int$ is the integral symbol.
* $f(x)$ is the integrand.
* $dx$ indicates that the integration is with respect to $x$.
* $F(x)$ is the anti-derivative of $f(x)$.
* $c$ is the constant of integration.
KEY TAKEAWAY: Don’t forget to add the constant of integration,
c, when finding indefinite integrals.
| Rule | Formula |
|---|---|
| Power Rule | $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$, $n \neq -1$ |
| Constant Multiple Rule | $\int kf(x) \, dx = k \int f(x) \, dx$ |
| Sum/Difference Rule | $\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx$ |
| Integral of $e^x$ | $\int e^x \, dx = e^x + c$ |
| Integral of $e^{kx}$ | $\int e^{kx} \, dx = \frac{1}{k}e^{kx} + c$ |
EXAM TIP: Practice applying these rules to various functions.
The definite integral of a function $f(x)$ over the interval $[a, b]$ is denoted by:
$$\int_a^b f(x) \, dx$$
where:
* $a$ is the lower limit of integration.
* $b$ is the upper limit of integration.
If $f$ is a continuous function on the interval $[a, b]$, and $F(x)$ is any anti-derivative of $f(x)$, then:
$$\int_a^b f(x) \, dx = F(b) - F(a)$$
This is often written as:
$$\int_a^b f(x) \, dx = [F(x)]_a^b$$
COMMON MISTAKE: Forgetting to evaluate the anti-derivative at both the upper and lower limits.
If $f(x)$ is integrable on an interval containing $a$, $b$, and $c$, then:
$$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx$$
$$\int_a^a f(x) \, dx = 0$$
$$\int_a^b kf(x) \, dx = k \int_a^b f(x) \, dx$$
$$\int_a^b [f(x) \pm g(x)] \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx$$
$$\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx$$
These properties can be combined to simplify complex integrals.
| Property | Formula |
|---|---|
| Interval Additivity | $\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx$ |
| Integral of Zero | $\int_a^a f(x) \, dx = 0$ |
| Constant Multiple | $\int_a^b kf(x) \, dx = k \int_a^b f(x) \, dx$ |
| Sum/Difference | $\int_a^b [f(x) \pm g(x)] \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx$ |
| Reversing Limits | $\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx$ |
Using Interval Additivity:
If $\int_1^5 f(x) \, dx = 10$ and $\int_1^3 f(x) \, dx = 4$, find $\int_3^5 f(x) \, dx$.
$$\int_1^5 f(x) \, dx = \int_1^3 f(x) \, dx + \int_3^5 f(x) \, dx$$
$\$10 = 4 + \int_3^5 f(x) \, dx$$
$$\int_3^5 f(x) \, dx = 6$$
Using Constant Multiple and Sum Rule:
If $\int_0^2 x \, dx = 2$ and $\int_0^2 x^2 \, dx = \frac{8}{3}$, find $\int_0^2 (3x - 2x^2) \, dx$.
$$\int_0^2 (3x - 2x^2) \, dx = 3\int_0^2 x \, dx - 2\int_0^2 x^2 \, dx$$
$$= 3(2) - 2(\frac{8}{3})$$
$$= 6 - \frac{16}{3} = \frac{2}{3}$$
STUDY HINT: Practice using these properties in combination to solve more complex problems.
If $f(x) \geq 0$ for all $x \in [a, b]$, then the area of the region between the curve $y = f(x)$, the x-axis, and the lines $x = a$ and $x = b$ is given by:
$$Area = \int_a^b f(x) \, dx$$
If $f(x) \leq 0$ for all $x \in [a,b]$, then:
$$Area = - \int_a^b f(x) \, dx$$
If $f(x)$ changes sign on $[a,b]$, then the area is found by breaking the integral into sections where $f(x)$ is either always positive or always negative.
The area of the region bounded by two curves $y = f(x)$ and $y = g(x)$ and the lines $x = a$ and $x = b$, where $f(x) \geq g(x)$ for all $x \in [a, b]$, is given by:
$$Area = \int_a^b [f(x) - g(x)] \, dx$$
The average value of a continuous function $f(x)$ on the interval $[a, b]$ is given by:
$$f_{avg} = \frac{1}{b - a} \int_a^b f(x) \, dx$$
Area under a curve:
Find the area under the curve $y = x^2$ from $x = 0$ to $x = 2$.
$$Area = \int_0^2 x^2 \, dx = [\frac{x^3}{3}]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}$$
Area between two curves:
Find the area between the curves $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$.
$$Area = \int_0^1 (x - x^2) \, dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = (\frac{1}{2} - \frac{1}{3}) - (0 - 0) = \frac{1}{6}$$
Average value of a function:
Find the average value of $f(x) = x^2$ on the interval $[1, 3]$.
$$f_{avg} = \frac{1}{3 - 1} \int_1^3 x^2 \, dx = \frac{1}{2} [\frac{x^3}{3}]_1^3 = \frac{1}{2} (\frac{27}{3} - \frac{1}{3}) = \frac{1}{2} (\frac{26}{3}) = \frac{13}{3}$$
VCAA FOCUS: VCAA exams often include questions that require you to find areas under curves and between curves, as well as average values of functions using definite integrals.
A function $f(x)$ is even if $f(-x) = f(x)$ for all $x$ in its domain. The graph of an even function is symmetric with respect to the y-axis.
For even functions:
$$\int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx$$
A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain. The graph of an odd function is symmetric with respect to the origin.
For odd functions:
$$\int_{-a}^a f(x) \, dx = 0$$
$f(x) = x^2$ is an even function.
$$\int_{-2}^2 x^2 \, dx = 2 \int_0^2 x^2 \, dx = 2[\frac{x^3}{3}]_0^2 = 2(\frac{8}{3}) = \frac{16}{3}$$
$f(x) = x^3$ is an odd function.
$$\int_{-2}^2 x^3 \, dx = 0$$
REMEMBER: Recognizing even and odd functions can significantly simplify definite integral calculations. If the limits are symmetric about the origin, and the function is odd, the integral is zero.
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