The Fundamental Theorem of Calculus and Accumulation Functions: AP Calculus Study Guide
Introduction
Greetings, future calculus champions! 🤓 Ready to dive into one of the grandest bridges in mathematics? No, not the Golden Gate Bridge—but the bridge between derivatives and integrals! We’re talking about the Fundamental Theorem of Calculus (FTC). It’s where the world of differentiation meets integration like jazz and classical music coming together to create a mathematical symphony. 🎺🎻
The Fundamental Theorem of Calculus (FTC)
So, what's this big deal about the FTC? Well, it’s like finding out that Batman and Bruce Wayne are the same person—mindblowing! It connects two seemingly different concepts: differentiation and integration. The FTC has two parts, and they go together like peanut butter and jelly. 🍞🥜🍓
FTC Part 1: The Relationship Revealed
Imagine you have a function ( f(t) ) that’s continuous on an interval ([a, b]). Now, define a new function ( F(x) ) that represents the area under the curve of ( f(t) ) from ( a ) to ( x ). Mathematically, that’s written as: [ F(x) = \int_{a}^{x} f(t) , dt ]
The FTC Part 1 states that ( F(x) ) has an astonishing superpower: it’s differentiable, and its derivative is just ( f(x) )! 📈💥 This means: [ \frac{d}{dx} \left[ \int_{a}^{x} f(t) , dt \right] = f(x) ]
It’s like saying: "Hey, if you differentiate the area under the curve from point ( a ) to ( x ), you just get your original function back!"
FTC Part 2: The AntiDerivative Makeover
FTC Part 2 is like the grand finale where the integral gets a complete makeover. If ( f ) is indeed the derivative of some function ( F ) (differentiable on [a, b]), then: [ \int_{a}^{b} f(x) , dx = F(b)  F(a) ]
In simpler terms, the definite integral of ( f ) from ( a ) to ( b ) is simply the net change in ( F ) over that interval. Mind = Blown. 🤯
Accumulation Functions and Examples
Alrighty, let’s get those brain gears turning with some examples! 🧠⚙️
Example 1: Basic Bound
You have a function ( g(x) ) defined as: [ g(x) = \int_{5}^{x} t^{1/4} , dt ]
To find ( g'(16) ), we use the FTC Part 1. According to the theorem, the derivative is simply the function inside the integral, evaluated at ( x ): [ g'(x) = x^{1/4} ]
Now substitute ( x = 16 ): [ g'(16) = 16^{1/4} = 2 ]
Not too difficult, right? 🎩✨
Example 2: Fancy Upper Bound
Consider ( F(x) ) given by: [ F(x) = \int_{3}^{x^2} (t+4) , dt ]
Here, ( x^2 ) is the upperbound function. Again, apply FTC Part 1 along with the chain rule: [ \frac{d}{dx} \left[ \int_{3}^{x^2} (t+4) , dt \right] ]
First, find the antiderivative evaluated at ( x^2 ): [ F'(x) = (x^2 + 4) \times \frac{d}{dx} x^2 ]
Then apply the chain rule: [ F'(x) = (x^2 + 4) \times 2x = 2x(x^2 + 4) ]
Bingo! You’ve got ( F'(x) = 2x(x^2 + 4) ). 🍽️🧩
Practice Problems
It’s your turn to shine! 🌟 Ready for some handson practice? Tackle these problems like a math superhero. 🦸♂️🦸♀️
 Given ( g(x) = \int_{0}^{x} \sqrt{8 + \cos(t)} , dt ), find ( g'(0) ).
 Given ( g(x) = \int_{1}^{x} (5t^2 + 2t) , dt ), find ( g'(3) ).
 Given ( g(x) = \int_{0}^{x} \sqrt{\sin(t) + 15} , dt ), find ( g'(\frac{\pi}{2}) ).
 Given ( F(x) = \int_{3x}^{1} \sec^2(t) , dt ), find ( F'(x) ).
Solutions

For ( g(x) = \int_{0}^{x} \sqrt{8 + \cos(t)} , dt ),
 By FTC Part 1, ( g'(x) = \sqrt{8 + \cos(x)} ).
 Thus, ( g'(0) = \sqrt{8 + \cos(0)} = \sqrt{8 + 1} = 3 ).

For ( g(x) = \int_{1}^{x} (5t^2 + 2t) , dt ),
 By FTC Part 1, ( g'(x) = 5x^2 + 2x ).
 Thus, ( g'(3) = 5(3)^2 + 2(3) = 51 ).

For ( g(x) = \int_{0}^{x} \sqrt{\sin(t) + 15} , dt ),
 By FTC Part 1, ( g'(x) = \sqrt{\sin(x) + 15} ).
 Thus, ( g'(\frac{\pi}{2}) = \sqrt{\sin(\frac{\pi}{2}) + 15} = 4 ).

For ( F(x) = \int_{3x}^{1} \sec^2(t) , dt ),
 Flip the bounds: ( F(x) = \int_{1}^{3x} \sec^2(t) , dt ).
 By FTC Part 1, ( F'(x) = \sec^2(3x) \times \frac{d}{dx}(3x) = \sec^2(3x) \times 3 = 3\sec^2(3x) ).
Conclusion
Nice work, mathlete! You've just unlocked the power of the Fundamental Theorem of Calculus. Whether the bounds are plain ( x ) or something more sophisticated like ( x^2 ), you've got the tools to tackle those integrals and derivatives. Practice makes perfect, so keep at it, and soon you'll be doing calculus with the ease of a seasoned pro. 📚✨
Remember, calculus is a journey, and every problem you solve gets you a step closer to mathematical mastery. Persevere, practice, and may the derivatives be ever in your favor! 💪📐