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:
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:
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:
This is often written as:
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:
\[\int_a^a f(x) \, dx = 0\]
\[\int_a^b kf(x) \, dx = k \int_a^b f(x) \, dx\]
\[\int_a^b [f(x) \pm g(x)] \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx\]
\[\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx\]
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 \(\int_3^5 f(x) \, dx\).
Using Constant Multiple and Sum Rule:
If \(\int_0^2 x \, dx = 2\) and \(\int_0^2 x^2 \, dx = \frac{8}{3}\), find \(\int_0^2 (3x - 2x^2) \, dx\).
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:
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:
The average value of a continuous function \(f(x)\) on the interval \([a, b]\) is given by:
Area under a curve:
Find the area under the curve \(y = x^2\) from \(x = 0\) to \(x = 2\).
Area between two curves:
Find the area between the curves \(y = x^2\) and \(y = x\) from \(x = 0\) to \(x = 1\).
Average value of a function:
Find the average value of \(f(x) = x^2\) on the interval \([1, 3]\).
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:
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:
\(f(x) = x^2\) 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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