Welcome to this week's class recap! In our most recent session covering Chapter 7, Sections 1-4, we dove deep into the mechanics of integration. From manipulating definite integrals to mastering the art of $u$-substitution, these techniques are the building blocks for the more complex calculus problems ahead.

Below is a breakdown of the key concepts from the lecture and the attached handwritten notes.

1. Linearity and Properties of Definite Integrals

One of the first things we reviewed was how to break down complex integrands using the properties of linearity. As seen in the class notes (Problem 53), when faced with an integral like:

$$\int_{-4}^{2} [f(x) + 2x + 5] dx$$

It is often easiest to split this into three manageable parts:

$$\int_{-4}^{2} f(x) dx + \int_{-4}^{2} 2x dx + \int_{-4}^{2} 5 dx$$

By evaluating the known area of $f(x)$ (which was given as a signed area of $-3$) and using the Power Rule for the polynomial terms, we arrived at the final answer of 15.

2. Trigonometric Integrals

Never forget your Pythagorean identities! In Problem 15, we encountered:

$$\int (2 + \tan^2 \theta) d\theta$$

At first glance, this might look like it requires a complex substitution. However, by rewriting the integrand using the identity $1 + \tan^2 \theta = \sec^2 \theta$, we can simplify the problem significantly:

$$\int (1 + [1 + \tan^2 \theta]) d\theta = \int (1 + \sec^2 \theta) d\theta$$

Since the integral of $\sec^2 \theta$ is simply $\tan \theta$, the solution flows naturally: $\theta + \tan \theta + C$.

3. Advanced U-Substitution Techniques

The bulk of Sections 7.3 and 7.4 focused on the versatility of Integration by Substitution. We covered three specific scenarios:

  • Splitting the Numerator: In Problem 45, we dealt with $\int \frac{1+x}{1+x^2} dx$. By splitting the fraction into $\frac{1}{1+x^2}$ and $\frac{x}{1+x^2}$, we solve the first part as an inverse tangent function ($\tan^{-1}x$) and the second part using a simple $u$-substitution ($u=1+x^2$).
  • Solving for X (Change of Variables): Sometimes, $du$ doesn't perfectly cancel out all $x$ terms. In Problem 47, $\int x(2x+5)^8 dx$, we set $u=2x+5$. To handle the remaining $x$, we rearranged the equation to solve for $x$ in terms of $u$: $x = \frac{u-5}{2}$. This allows us to rewrite the entire integral in terms of $u$.
  • Changing Limits for Definite Integrals: When performing substitution on definite integrals, remember to change your limits of integration!
    For example, in Problem 67 ($\int_1^2 x\sqrt{x-1} dx$), by setting $u=x-1$, our lower limit changes from $x=1$ to $u=0$, and the upper limit from $x=2$ to $u=1$. This saves you from having to back-substitute at the very end.

Class Resources

To help you practice these concepts, please review the PowerPoints and the handwritten problem solutions attached to this post.

  • Lecture Slides: Sections 7-1 & 7-2, Sections 7-3 & 7-4
  • Handwritten Solutions: See attached PDF for step-by-step derivations.

If you have questions regarding these sections or the previous week's lessons, please fill out the Questions Form so we can address them at the start of next week's class. Keep practicing those integrals!