Welcome to Calculus 2 (Spring 2022)! I am thrilled to have you in class. This website will serve as your central hub for all course materials, including PowerPoints, review sheets, and class updates. Below you will find the critical information to get started, along with a recap of our first lesson on Applications of Integration.

📝 Class Logistics & To-Do List

Please ensure you complete the following items as soon as possible to stay on track:

  • Review the Syllabus: Understand the grading policy (Tests 15%, Final 30%) and attendance requirements.
  • Register for WebAssign: This is required for your homework.
    • Class Code: trcc.mohegan 1965 6769
  • Fill out the Contact Form: Please complete the Contact Form linked on the main site. This is crucial for me to reach you regarding class cancellations or urgent updates.

📚 Lesson Summary: Areas Between Curves (Section 5.1)

Tonight we jumped right into Chapter 5: Applications of Integration. Specifically, we looked at how to calculate the area of a region bounded by two curves. In Calculus 1, you learned that the integral represents the area under a curve. Now, we extend that concept to find the area between curves.

1. Integrating with respect to $x$ (Top - Bottom)

If a region $S$ is bounded above by $y=f(x)$ and bounded below by $y=g(x)$ between $x=a$ and $x=b$, the area is defined as:

$$A = \int_{a}^{b} [f(x) - g(x)] \, dx$$

Key Strategy: Always sketch the graph first! You need to visually identify which function is the "Top" and which is the "Bottom."

Example from class notes: For the region bounded by $y = e^x$ and $y = x$ from $x=0$ to $x=1$, we noticed $e^x$ is the top curve. The setup is:

$$ \int_{0}^{1} (e^x - x) \, dx = \left[ e^x - \frac{1}{2}x^2 \right]_0^1 = e - 1.5 $$

2. Integrating with respect to $y$ (Right - Left)

Sometimes, it is mathematically easier to treat $x$ as a function of $y$ ($x = f(y)$). This is usually necessary when curves overlap in a way that makes vertical slicing difficult. In this case, we subtract the Left curve from the Right curve:

$$A = \int_{c}^{d} [x_R - x_L] \, dy$$

Example from class notes: We solved for the area enclosed by the line $y = x - 1$ and the parabola $y^2 = 2x + 6$. By solving for $x$, we get $x_R = y + 1$ (the line) and $x_L = \frac{1}{2}y^2 - 3$ (the parabola). Integrating from $y=-2$ to $y=4$ yielded an area of 18.

🎥 Review Materials

I have attached the PDFs for the lecture slides and the handwritten notes from our session. Please review the examples involving sine and cosine curves ($y=\sin x$ and $y=\cos x$), as identifying the intersection points is vital for setting up your integration limits correctly.

Office Hours: Tuesdays & Thursdays, 7:00pm - 8:00pm (by appointment).
Let's have a great semester!