BackBusiness Calculus II: Integration, Applications, and Average Value Study Guide
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Integration Techniques and Applications
Integration by Parts
Integration by parts is a fundamental technique for evaluating integrals where the standard methods do not apply directly. It is based on the product rule for differentiation and is given by:
Formula: $\int u\,dv = uv - \int v\,du$
Key Steps:
Identify parts: Let $u$ be a function that becomes simpler when differentiated, and $dv$ be a function that is easy to integrate.
Differentiate $u$ to get $du$; integrate $dv$ to get $v$.
Substitute into the formula and simplify.
Example: $\int x e^{2x} dx$
Let $u = x$, $dv = e^{2x} dx$
$du = dx$, $v = \frac{1}{2}e^{2x}$
So, $\int x e^{2x} dx = x \cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x} dx = \frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C$
Integration by Substitution
Substitution is used to simplify integrals by changing variables, often making the integral easier to solve.
Formula: If $u = g(x)$, then $\int f(g(x))g'(x)dx = \int f(u)du$
Key Steps:
Let $u$ be a function inside the integrand.
Compute $du$ and substitute all $x$ terms.
Integrate with respect to $u$.
Example: $\int x \sqrt{x^2 + 1} dx$
Let $u = x^2 + 1$, $du = 2x dx$
$x dx = \frac{1}{2} du$
So, $\int x \sqrt{x^2 + 1} dx = \int \sqrt{u} \cdot \frac{1}{2} du = \frac{1}{2} \int u^{1/2} du = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3}(x^2 + 1)^{3/2} + C$
Definite Integrals and Area Under Curves
Definite Integrals
Definite integrals are used to compute the net area under a curve between two points. The result is a number, not a function.
Formula: $\int_a^b f(x) dx$
Key Properties:
If $f(x)$ is above the $x$-axis, the area is positive; if below, negative.
To find the total area (regardless of sign), take the absolute value of each region.
Example: Find the area under $f(x) = x^3 - x$ from $x = -1$ to $x = 2$.
Set up the integral: $\int_{-1}^2 (x^3 - x) dx$
Evaluate: $\left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^2$
Area Between Curves
To find the area between two curves, subtract the lower function from the upper function and integrate over the interval where they intersect.
Formula: $\int_a^b [f(x) - g(x)] dx$ where $f(x)$ is the upper curve and $g(x)$ is the lower curve.
Example: Area between $f(x) = 3 - x^2$ and $g(x) = x$ from $x = -1$ to $x = 2$:
Set up: $\int_{-1}^2 [(3 - x^2) - x] dx$
Evaluate: $\left[3x - \frac{x^3}{3} - \frac{x^2}{2}\right]_{-1}^2$
Average Value of a Function
Average Value Formula
The average value of a continuous function $f(x)$ over the interval $[a, b]$ is given by:
Formula: $f_{avg} = \frac{1}{b-a} \int_a^b f(x) dx$
Key Steps:
Set up the definite integral over the interval.
Divide the result by the length of the interval $(b-a)$.
Example: Find the average value of $f(x) = x$ over $[1, 3]$:
$f_{avg} = \frac{1}{3-1} \int_1^3 x dx = \frac{1}{2} \left[\frac{x^2}{2}\right]_1^3 = \frac{1}{2} \left(\frac{9}{2} - \frac{1}{2}\right) = \frac{1}{2} \cdot 4 = 2$
Applications: Volume of Revolution
Volume Using the Disk/Washer Method
To find the volume of a solid formed by revolving a region around the $x$-axis, use the disk or washer method.
Formula: $V = \pi \int_a^b [R(x)^2 - r(x)^2] dx$ where $R(x)$ is the outer radius and $r(x)$ is the inner radius.
Example: Volume generated by revolving $f(x) = 9 - x^2$ over $[1, 2]$:
$V = \pi \int_1^2 (9 - x^2)^2 dx$
Expand and integrate: $\pi \int_1^2 (81 - 18x^2 + x^4) dx$
Applications: Physics and Average Acceleration
Definite Integral for Net Change
The definite integral of a rate function (such as acceleration) over an interval gives the net change in the corresponding quantity (such as velocity).
Formula: $\int_a^b a(t) dt = v(b) - v(a)$
Example: If $a(t) = 50/(t+1)^2$, the net change in velocity from $t=1$ to $t=47$ is:
$\int_1^{47} \frac{50}{(t+1)^2} dt = 50 \left[ -\frac{1}{t+1} \right]_1^{47} = 50 \left( -\frac{1}{48} + \frac{1}{2} \right) = 50 \left( \frac{23}{24} \right) = 47.92$
Average Acceleration
The average value of acceleration over an interval is found using the average value formula.
Formula: $a_{avg} = \frac{1}{b-a} \int_a^b a(t) dt$
Example: For $a(t) = 50/(t+1)^2$ over $[1, 47]$:
$a_{avg} = \frac{1}{47-1} \int_1^{47} \frac{50}{(t+1)^2} dt = \frac{47.92}{46} = 1.042$
Summary Table: Key Integration Techniques and Applications
Technique/Application | Formula | Example |
|---|---|---|
Integration by Parts | $\int u\,dv = uv - \int v\,du$ | $\int x e^{2x} dx$ |
Substitution | $\int f(g(x))g'(x)dx = \int f(u)du$ | $\int x \sqrt{x^2 + 1} dx$ |
Definite Integral (Area) | $\int_a^b f(x) dx$ | Area under $f(x)$ from $a$ to $b$ |
Area Between Curves | $\int_a^b [f(x) - g(x)] dx$ | Area between $f(x)$ and $g(x)$ |
Average Value | $f_{avg} = \frac{1}{b-a} \int_a^b f(x) dx$ | Average value of $f(x)$ over $[a, b]$ |
Volume of Revolution | $V = \pi \int_a^b [R(x)^2 - r(x)^2] dx$ | Volume generated by revolving a region |
Additional info: These notes cover advanced integration techniques, applications to area and volume, and average value concepts, all of which are central to Business Calculus II and its applications in business, economics, and science.
