Welcome to the Spring Semester! It was great to see everyone for our first night of class. We have an exciting semester ahead in MAT K256 Calculus II. Below you will find a summary of the administrative details we discussed, as well as a recap of our first lesson on Finding Areas Between Curves.

Part 1: Course Logistics & Syllabus Highlights

Before jumping into the math, we reviewed the structure of the course. Here are the key takeaways from the syllabus to keep you on track:

  • Grading Breakdown: Your grade is calculated based on Participation (10%), 4 Tests (15% each), and the Final Exam (30%).
  • Attendance: Attendance is crucial for your success. Missing class means missing participation points and vital material.
  • Office Hours: I am available Tuesdays and Thursdays from 11 am to 4 pm (by appointment only). Please email me at tbaker@trcc.commnet.edu to schedule time if you need extra help.
  • Textbook: We are using Calculus Early Transcendentals.

Part 2: Lesson Recap - Section 6.1 (Areas Between Curves)

We started our journey into integration applications by looking at how to find the area of a region $S$ that lies between two curves. In Calculus I, you learned how to find the area under a single curve to the x-axis. Now, we are finding the area trapped between two functions, $f(x)$ and $g(x)$.

Case 1: Vertical Slicing (Integrating with respect to $x$)

When the region is bounded by a top curve $y = f(x)$ and a bottom curve $y = g(x)$ between vertical lines $x=a$ and $x=b$, we calculate the area by subtracting the bottom function from the top function:

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

Example from Class:
We looked at the region enclosed by the parabolas $y = x^2$ and $y = 2x - x^2$.

  1. Find Intersections: First, we set the equations equal to find our bounds ($a$ and $b$). $$x^2 = 2x - x^2 \implies 2x^2 - 2x = 0 \implies 2x(x-1) = 0$$ This gives us intersection points at $x=0$ and $x=1$.
  2. Set up the Integral: In the interval $(0, 1)$, $2x - x^2$ is the upper curve. $$A = \int_{0}^{1} [(2x - x^2) - (x^2)] \, dx = \int_{0}^{1} (2x - 2x^2) \, dx$$
  3. Evaluate: $$\left[ x^2 - \frac{2x^3}{3} \right]_0^1 = \left(1 - \frac{2}{3}\right) - 0 = \frac{1}{3}$$

Case 2: Horizontal Slicing (Integrating with respect to $y$)

Sometimes, it is easier to treat $x$ as a function of $y$. If a region is bounded by a "Right" curve $x_R$ and a "Left" curve $x_L$, we integrate with respect to $y$ from $c$ to $d$.

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

We practiced this with the area enclosed by the line $y = x - 1$ and the parabola $y^2 = 2x + 6$. By rewriting these as $x = y+1$ (Right) and $x = \frac{1}{2}y^2 - 3$ (Left), we were able to set up a simpler integral over the $y$-interval $[-2, 4]$, resulting in a total area of 18.

Looking Ahead

Please make sure to review the attached class notes for more examples, including the trig functions example ($y=\sin x$ and $y=\cos x$) and the substitution method we used for the radical function. Next class (Thursday, 1/26), we will move on to Sections 6.2 and 6.3 regarding Volumes. See you there!