AP Calculus AB/BC: Integrals — Fundamental Theorem, U-Substitution, Integration by Parts (BC)

TL;DR

Integrals help you find the accumulation of a rate of change, often representing area under a curve. The Fundamental Theorem of Calculus links derivatives and integrals, providing a way to evaluate definite integrals. U-substitution simplifies integrals by transforming them, while integration by parts (BC topic) handles product functions.

1. The Mental Model

Think of integration as going backward from a rate of change to the total amount that has accumulated. If you know how fast something is changing, integration tells you how much of that "something" you have now. It's like finding the total distance traveled if you know your speed at every moment.

2. The Core Material

Integrals are essentially fancy summators. A definite integral ($\int_a^b f(x) \,dx$) calculates the exact accumulation of a function $f(x)$ from point $a$ to point $b$. An indefinite integral ($\int f(x) \,dx$) finds the family of functions whose derivative is $f(x)$, always including an arbitrary constant $C$.

The Fundamental Theorem of Calculus (FTC)

The FTC Part 1 states that if $F(x) = \int_a^x f(t) \,dt$, then $F'(x) = f(x)$. This means differentiation and integration are inverse operations.

The FTC Part 2 is what you'll use most often for definite integrals:
$\int_a^b f(x) \,dx = F(b) - F(a)$, where $F(x)$ is any antiderivative of $f(x)$ (meaning $F'(x) = f(x)$).

To use FTC Part 2, you:
1. Find the antiderivative $F(x)$ of $f(x)$.
2. Evaluate $F(b)$.
3. Evaluate $F(a)$.
4. Subtract $F(a)$ from $F(b)$.

Example: $\int_1^3 2x \,dx$
1. Antiderivative of $2x$ is $x^2$. So $F(x) = x^2$.
2. $F(3) = 3^2 = 9$.
3. $F(1) = 1^2 = 1$.
4. $F(3) - F(1) = 9 - 1 = 8$.

U-Substitution (The "Reverse Chain Rule")

U-substitution helps integrate functions that look like they came from a chain rule derivative. The goal is to simplify the integrand by replacing a complex inner function ($u$) and its derivative ($du$).

Steps:
1. Choose a part of the integrand to be $u$, usually the "inside" function of a composition.
2. Find $du$ by differentiating $u$ with respect to the original variable (e.g., $dx$).
3. Rewrite the integral entirely in terms of $u$ and $du$. You might need to adjust constants.
4. Integrate with respect to $u$.
5. Substitute the original expression back in for $u$.
6. For definite integrals, either change the limits of integration to be in terms of $u$ or substitute back to the original variable before evaluating.

Example: $\int x \cos(x^2) \,dx$
1. Let $u = x^2$.
2. Then $du = 2x \,dx$.
3. We have $x \,dx$ in the integral, so $x \,dx = \frac{1}{2} du$.
4. Rewrite: $\int \cos(u) \cdot \frac{1}{2} \,du = \frac{1}{2} \int \cos(u) \,du$.
5. Integrate: $\frac{1}{2} \sin(u) + C$.
6. Substitute back: $\frac{1}{2} \sin(x^2) + C$.

graph TD
    A["Identify Complex Integral"] --> B["Look for composite function (inner function)"];
    B --> C{"Is there an 'inner' function and its derivative (or a scalar multiple)?"};
    C -- Yes --> D["Choose u = inner function"];
    D --> E["Calculate du = u' dx"];
    E --> F["Substitute u and du into integral"];
    F --> G["Simplify and integrate with respect to u"];
    G --> H["Substitute original expression back for u"];
    H --> I{"Definite Integral?"};
    I -- Yes --> J["Change limits to u-values OR substitute back before evaluating"];
    I -- No --> K["Add + C"];

Integration by Parts (BC Only)

Integration by parts is used when you have a product of two functions that isn't easily handled by u-substitution. It's based on the product rule for derivatives: $\int u \,dv = uv - \int v \,du$. This formula helps transform a difficult integral into a potentially easier one.

The trick is choosing which part of the integrand is $u$ and which is $dv$. A common mnemonic for choosing $u$ is LIATE:
- Logarithmic functions ($\ln x$)
- Inverse trigonometric functions ($\arctan x$)
- Algebraic functions ($x^n$)
- Trigonometric functions ($\sin x$, $\cos x$)
- Exponential functions ($e^x$)

You usually pick $u$ as the function that comes first in LIATE, and $dv$ as the remaining part. You want $u$ to become simpler when derived, and $dv$ to be easily integrable.

Steps:
1. Choose $u$ and $dv$.
2. Find $du$ by differentiating $u$.
3. Find $v$ by integrating $dv$.
4. Apply the formula: $\int u \,dv = uv - \int v \,du$.
5. Evaluate the new, often simpler, integral $\int v \,du$. You might need to apply integration by parts again!

Example: $\int x e^x \,dx$
1. Using LIATE, $x$ is Algebraic (A) and $e^x$ is Exponential (E). So, let $u = x$ and $dv = e^x \,dx$.
2. Differentiate $u$: $du = 1 \,dx$.
3. Integrate $dv$: $v = \int e^x \,dx = e^x$.
4. Apply formula: $\int x e^x \,dx = x e^x - \int e^x \,dx$.
5. Evaluate the new integral: $\int e^x \,dx = e^x$.
6. Final answer: $x e^x - e^x + C$.

3. Worked Example

Let's evaluate $\int_0^{\pi/2} x \sin(x) \,dx$ using integration by parts (and then FTC).

This is a product of two functions, $x$ and $\sin(x)$.
Using LIATE: $x$ is Algebraic, $\sin(x)$ is Trigonometric. So, choose $u=x$.

1. Choose $u$ and $dv$:
   Let $u = x$
   Let $dv = \sin(x) \,dx$

2. Find $du$ and $v$:
   Differentiate $u$: $du = 1 \,dx$
   Integrate $dv$: $v = \int \sin(x) \,dx = -\cos(x)$ (no +C for $v$ here, it gets handled later)

3. Apply the Integration by Parts formula:
   $\int u \,dv = uv - \int v \,du$
   $\int x \sin(x) \,dx = x(-\cos(x)) - \int (-\cos(x))(1 \,dx)$
   $\int x \sin(x) \,dx = -x \cos(x) + \int \cos(x) \,dx$

4. Evaluate the new integral:
   $\int \cos(x) \,dx = \sin(x)$

5. Combine for the indefinite integral:
   $\int x \sin(x) \,dx = -x \cos(x) + \sin(x) + C$

6. Now, apply FTC Part 2 for the definite integral:
   $\int_0^{\pi/2} x \sin(x) \,dx = [-x \cos(x) + \sin(x)]_0^{\pi/2}$
   $= [(-\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})) - (-(0) \cos(0) + \sin(0))]$
   $= [(-\frac{\pi}{2} \cdot 0 + 1) - (0 \cdot 1 + 0)]$
   $= [0 + 1] - [0 + 0]$
   $= 1 - 0 = 1$

The value of the definite integral is $1$.

4. Key Takeaways

• The Fundamental Theorem of Calculus connects differentiation and integration, letting you evaluate definite integrals by finding antiderivatives.
• U-substitution simplifies integrals that look like they resulted from the chain rule.
• Always remember to differentiate your chosen $u$ to find $du$ for u-substitution.
• For definite integrals with u-substitution, either change the limits or substitute back before evaluating.
• Integration by parts is for product functions and uses the mnemonic LIATE to help choose $u$.
• When using integration by parts, you want $u$ to simplify upon differentiation and $dv$ to be easy to integrate.
• Integration is about accumulation; it finds the "total" given a "rate."

Common Mistakes to Avoid:
- Forgetting the "+ C" for indefinite integrals.
- Not changing the limits of integration when doing definite integrals with u-substitution.
- Incorrectly choosing $u$ and $dv$ for integration by parts, leading to a more complex integral.
- Distributing the negative sign incorrectly when applying $F(b) - F(a)$ in FTC.

5. Now Try It

Evaluate the definite integral $\int_0^1 x^2 e^{x^3} \,dx$. Set a timer for 15 minutes.
What to do: Decide whether you need FTC, u-substitution, or integration by parts. Carry out the steps carefully.
What success looks like: You should arrive at a single numerical answer, $ \frac{1}{3}(e-1) $.