Welcome back to Professor Baker's Math Class! In this lesson, we are bridging the gap between derivatives and integrals. We move from the concept of area under a curve to finding general antiderivatives, and then we learn one of the most powerful techniques in a calculus student's toolkit: Substitution.

Section 5-4: Indefinite Integrals and the Net Change Theorem

First, it is crucial to understand the distinction in notation and result between definite and indefinite integrals, as highlighted in the class notes:

  • Definite Integral: $\int_{a}^{b} f(x) dx$ results in a number (representing net area).
  • Indefinite Integral: $\int f(x) dx$ results in a function (a family of antiderivatives).

Don't forget the arbitrary constant $+ C$ when dealing with indefinite integrals! For example, using the standard integration rules provided in the notes:

$$\int (10x^4 - 2\sec^2 x) dx = 2x^5 - 2\tan x + C$$

Application: Particle Motion

One of the most important applications of integration is finding displacement and distance from velocity. As we saw in the class notes with the function $v(t) = t^2 - t - 6$, there is a big difference between the two:

  • Displacement: This is simply the definite integral of velocity, $\int_{t_1}^{t_2} v(t) dt$. It tells you how far you are from the starting point.
  • Total Distance Traveled: This is the integral of the absolute value of velocity, $\int_{t_1}^{t_2} |v(t)| dt$.

To find the total distance, you must determine where the particle stops (where $v(t)=0$) and split the integral. In our example on the interval $[1, 4]$, the particle stops at $t=3$. We had to calculate the area from 1 to 3 and add the absolute value of the area from 3 to 4. This changed our result from a displacement of $-4.5$ to a total distance of $61/6$.

Section 5-5: The Substitution Rule

Section 5-5 introduces the Substitution Rule (often called U-Substitution), which can be thought of as the "Reverse Chain Rule." This method helps us integrate composite functions.

The Strategy: Look for a function inside the integral, let's call it $u$, such that its derivative $du$ is also present in the integrand (potentially differing by a constant multiplier).

Example Breakdown

Let's look at a classic example from the notes:

$$\int x^3 \cos(x^4 + 2) dx$$
  1. Choose $u$: We see an "inner" function inside the cosine. Let $u = x^4 + 2$.
  2. Differentiate: Find $du$. $du = 4x^3 dx$.
  3. Match terms: We have $x^3 dx$ in our integral, but our $du$ has a 4. We can adjust this: $\frac{1}{4} du = x^3 dx$.
  4. Substitute: Replace terms to switch entirely to $u$. $$\int \cos(u) \cdot \frac{1}{4} du = \frac{1}{4} \int \cos(u) du$$
  5. Integrate: The integral of $\cos(u)$ is $\sin(u)$. $$= \frac{1}{4} \sin(u) + C$$
  6. Back-Substitute: Replace $u$ with the original $x$ terms. $$= \frac{1}{4} \sin(x^4 + 2) + C$$

This method works for radicals, exponentials, and trigonometric functions alike. For example, when integrating $\int \tan x dx$, we rewrite it as $\int \frac{\sin x}{\cos x} dx$ and use $u = \cos x$ to find that the result involves the natural log: $-\ln|\cos x| + C$.

Be sure to download the attached PDF notes for the full table of integration formulas and detailed step-by-step solutions for these examples!