MAT 266 - Test 2 Prep

Enhanced Study Guide

6.6 Improper Integrals

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Core Idea:

Integrals over an infinite interval or with an infinite discontinuity. The key is to replace the "problem" part with a variable and then take a limit.

💡 Always Remember

An improper integral is just a definite integral wrapped in a limit. Solve the integral first, then evaluate the limit.

Key Formulas & Theorems:

Type 1: Infinite Intervals
If $\int_a^t f(x)dx$ exists for every $t \ge a$, then: $$ \int_a^\infty f(x)dx = \lim_{t \to \infty} \int_a^t f(x)dx $$ The integral converges if the limit is a finite number. Otherwise, it diverges.
Type 2: Discontinuous Integrands
If $f$ is continuous on $[a, b)$ and discontinuous at $b$, then: $$ \int_a^b f(x)dx = \lim_{t \to b^-} \int_a^t f(x)dx $$

Common Pitfalls

  • Forgetting to use limit notation. You must write $\lim_{t \to \infty}$.
  • Incorrectly handling integrals from $-\infty$ to $\infty$. You must split it into two separate improper integrals: $\int_{-\infty}^c f(x)dx + \int_c^\infty f(x)dx$. Both must converge for the original to converge.
  • Ignoring a vertical asymptote in the middle of an interval, like $\int_{-1}^1 \frac{1}{x} dx$. This must also be split.

Mathematician's Vocabulary

  • "The integral converges to a finite value, L."
  • "The integral diverges to infinity."
  • "We must investigate the behavior of the integrand at the point of discontinuity."
  • "The p-integral $\int_1^\infty \frac{1}{x^p}dx$ is a canonical example of convergence for $p > 1$."
  • "The value of the integral is well-defined if and only if the limit exists."
  • "By the Comparison Test, since the integrand is dominated by a convergent function, this integral must also converge."
  • "The singularity at $x=0$ requires us to split the domain of integration."
  • "The oscillatory nature of the sine function causes this integral to diverge."
  • "We can establish convergence without finding the exact value."
  • "This is a trivial case of divergence as the limit of the integrand is non-zero."

7.1 Area Between Curves

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Core Idea:

Find the area of a region bounded by two or more functions by integrating the difference between the "upper" function and the "lower" function.

💡 Always Remember

Area is always positive. The integral $\int (f-g)dx$ represents "net area". To get total geometric area, ensure you are integrating (Top - Bottom) or (Right - Left).

Key Formulas & Theorems:

Integrating with respect to x: $$ A = \int_a^b [f(x) - g(x)] dx \quad \text{where } f(x) \ge g(x) $$ (Top function minus bottom function)
Integrating with respect to y: $$ A = \int_c^d [f(y) - g(y)] dy \quad \text{where } f(y) \ge g(y) $$ (Right function minus left function)

Common Pitfalls

  • Not finding all points of intersection, leading to incorrect limits of integration.
  • Mixing up the top and bottom functions, resulting in a negative answer.
  • Forgetting to solve for $x$ in terms of $y$ when integrating with respect to $y$.
  • Failing to split the integral if the top/bottom function changes within the region.

Mathematician's Vocabulary

  • "We define a representative rectangle of height $(f(x)-g(x))$ and width $dx$."
  • "The area is the limit of the Riemann sum of these rectangular elements."
  • "To find the bounds of integration, we determine the intersection points of the curves."
  • "It is more convenient to integrate with respect to y, treating x as a function of y."
  • "The region is bounded above by the parabola and below by the line."
  • "We must partition the interval because the upper boundary is piecewise."
  • "The integrand represents the differential element of area, $dA$."
  • "By symmetry, we can calculate the area in the first quadrant and double it."
  • "Let's verify which function is dominant over the interval $[a, b]$."
  • "The choice of integration variable is dictated by the geometry of the region."

7.2 & 7.3 Volumes of Revolution

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Core Idea:

Find the volume of a 3D solid created by rotating a 2D region around an axis. The method (Disk/Washer vs. Shell) depends on how you slice the region relative to the axis of rotation.

💡 Always Remember

The radius is ALWAYS measured from the axis of rotation. If rotating around $x=c$, the radius is not just $x$, it's $|x-c|$.

Key Formulas & Theorems:

Disk/Washer Method (Slicing perpendicular to axis of rotation)
Rotation around x-axis: $V = \int_a^b \pi [R(x)^2 - r(x)^2] dx$
Rotation around y-axis: $V = \int_c^d \pi [R(y)^2 - r(y)^2] dy$
($R$ is outer radius, $r$ is inner radius. If solid, $r=0$).
Shell Method (Slicing parallel to axis of rotation)
Rotation around y-axis: $V = \int_a^b 2\pi r(x) h(x) dx$
Rotation around x-axis: $V = \int_c^d 2\pi r(y) h(y) dy$
($r$ is shell radius, $h$ is shell height).

Common Pitfalls

  • Forgetting to square the radii in the Washer method ($R^2$ not $R$).
  • Mixing up the radius and height in the Shell method.
  • Using the wrong method for the orientation (e.g., using $dx$ for a horizontal axis rotation with Shells).
  • Incorrectly calculating the radii when rotating around a line that is not the x or y-axis. Always (Top - Axis) or (Axis - Bottom).

Mathematician's Vocabulary

  • "The volume element, $dV$, for the washer method is $\pi(R^2-r^2)dx$."
  • "We integrate the cross-sectional area perpendicular to the axis of rotation."
  • "The shell method is preferable here to avoid solving for x."
  • "The radius of the cylindrical shell is the distance from the axis of revolution."
  • "The resulting solid is a torus."
  • "This solid of revolution has a cavity, necessitating the washer method."
  • "The height of the shell is given by the difference of the two functions."
  • "Summing the volumes of the infinitesimal shells gives the total volume."
  • "The choice of method depends on the orientation of the representative rectangle."
  • "The limits of integration are determined by the projection of the region onto the x-axis."

7.4 Arc Length

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Core Idea:

Calculates the exact length of a curve over an interval by integrating the length of infinitesimal hypotenuses of right triangles along the curve.

💡 Always Remember

Look for algebraic simplification under the square root! Most test problems are designed so that $1 + [f'(x)]^2$ becomes a perfect square.

Key Formulas & Theorems:

For a function $y=f(x)$ from $x=a$ to $x=b$: $$ L = \int_a^b \sqrt{1 + [f'(x)]^2} dx $$
For a function $x=g(y)$ from $y=c$ to $y=d$: $$ L = \int_c^d \sqrt{1 + [g'(y)]^2} dy $$

Common Pitfalls

  • Forgetting to square the derivative inside the square root.
  • Incorrectly simplifying the algebra under the square root.
  • Trying to integrate $\sqrt{1} + \sqrt{[f'(x)]^2}$, which is wrong. You cannot distribute a square root over addition.
  • Using the wrong limits of integration (e.g., y-values for a dx integral).

Mathematician's Vocabulary

  • "The arc length is the integral of the magnitude of the velocity vector."
  • "This integral gives the length of the rectified curve."
  • "The differential element of arc length, $ds$, is given by $\sqrt{1+(dy/dx)^2}dx$."
  • "The integrand must be non-negative, which is guaranteed by the square."
  • "This function is continuously differentiable on the interval, so its arc length is well-defined."
  • "The resulting integral is non-trivial and may not have an elementary antiderivative."
  • "The expression under the radical simplifies to a perfect square."
  • "This is a standard application of the Pythagorean theorem on an infinitesimal scale."
  • "We are parameterizing the curve with respect to x."
  • "The smoothness of the function is a necessary condition for this formula."

7.6 Work

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Core Idea:

Work is calculated by integrating force over distance. For problems like pumping liquid or lifting ropes, the force changes as the position changes.

💡 Always Remember

Draw a diagram and establish a coordinate system FIRST. The most common error is mixing up the expression for the distance a slice must be moved.

Key Formulas & Theorems:

General Formula: $$ W = \int_a^b F(x) dx $$
Pumping Liquids: $$ W = \int_a^b (\text{density}) \cdot (\text{Area of slice}) \cdot (\text{distance moved}) \cdot dy $$ Remember: Work = (weight of a layer) * (distance that layer is moved).
Springs (Hooke's Law): $F = kx$. $$ W = \int_a^b kx dx $$ First, find the spring constant $k$.

Common Pitfalls

  • Confusing weight-density (like 62.5 lb/ft³) with mass-density (like 1000 kg/m³). Mass-density must be multiplied by gravity ($g=9.8$).
  • Incorrectly defining the distance. If your slice is at position $y$ and the top is at $H$, the distance is $(H-y)$, not just $y$.
  • Using the wrong formula for the area of a slice (e.g., $\pi r^2$ for a square cross-section).
  • For rope problems, defining the weight incorrectly. The weight of the remaining rope depends on how much has been pulled up.

Mathematician's Vocabulary

  • "The work done is the line integral of the force field over the path."
  • "We consider an infinitesimal slice of water of volume $dV$."
  • "The force on this slice is its weight, which is density times volume times gravity."
  • "The work required to lift this slice is $dW = F \cdot d$."
  • "We sum the work for all slices by integrating over the fluid's vertical extent."
  • "Hooke's Law states that the restorative force is linear with respect to displacement."
  • "First, we must find the spring constant, k, from the given conditions."
  • "The coordinate system is chosen to simplify the expression for the distance."
  • "The geometry of the tank dictates the formula for the cross-sectional area."
  • "The total work is the net energy required to move the object against the force."