AP Calculus BC 6.4 The Fundamental Theorem of Calculus and Accumulation Functions Study Notes - New Syllabus
AP Calculus BC 6.4 The Fundamental Theorem of Calculus and Accumulation Functions Study Notes- New syllabus
AP Calculus BC 6.4 The Fundamental Theorem of Calculus and Accumulation Functions Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- The Fundamental Theorem of Calculus connects differentiation and integration.
Key Concepts:
- The Fundamental Theorem of Calculus and Accumulation Functions
The Fundamental Theorem of Calculus and Accumulation Functions
Accumulation Functions
An accumulation function is a function defined as the definite integral of another function (typically a rate function) from a fixed point \( a \) to a variable upper limit \( x \).
The general form is:
\( F(x) =\displaystyle \int_a^x f(t)\,dt \)
Here, \( F(x) \) represents the accumulated area under the curve of \( f(t) \) from \( t = a \) to \( t = x \). This function captures how much “total” of something has been accumulated up to a certain point \( x \).
Key ideas:
- The lower limit is fixed (usually a constant like 0 or 1).
- The upper limit is a variable (usually \( x \)), making \( F(x) \) a function.
- If \( f(t) \) is positive, \( F(x) \) increases; if \( f(t) \) is negative, \( F(x) \) decreases.
Example:
Let \( f(t) = t^2 \). Define \( F(x) =\displaystyle \int_0^x t^2 \, dt \). Find \( F(2) \) and \( F(3) \).
▶️Answer/Explanation
\( F(2) =\displaystyle \int_0^2 t^2 \, dt = \left[\dfrac{t^3}{3}\right]_0^2 = \dfrac{8}{3} \)
\( F(3) =\displaystyle \int_0^3 t^2 \, dt = \left[\dfrac{t^3}{3}\right]_0^3 = \dfrac{27}{3} = 9 \)
The Fundamental Theorem of Calculus (Part 1)
The Fundamental Theorem of Calculus Part 1 states that if \( f \) is continuous on \([a, b]\), and \( F(x) =\displaystyle \int_a^x f(t)\,dt \), then:
\( F'(x) = f(x) \)
Interpretation: The derivative of the accumulation function gives back the original function. In other words, differentiating an integral gives the integrand back, assuming the upper limit is just \( x \).
This establishes a deep connection between derivatives and integrals.
Example:
Let \( F(x) =\displaystyle \int_1^x \cos(t)\,dt \). Find \( F'(x) \).
▶️Answer/Explanation
By FTC Part 1: \( F'(x) = \cos(x) \)
The Fundamental Theorem of Calculus (Part 2)
The Fundamental Theorem of Calculus Part 2 states that if \( F \) is any antiderivative of \( f \) on \([a, b]\), then:
\(\displaystyle \int_a^b f(x)\,dx = F(b) – F(a) \)
Interpretation: The definite integral of \( f(x) \) from \( a \) to \( b \) is the net change in any antiderivative \( F \) over that interval.
It allows us to evaluate a definite integral without calculating area directly — just use any antiderivative.
Example:
Evaluate \(\displaystyle \int_2^5 (3x^2 + 4x)\,dx \).
▶️Answer/Explanation
Find antiderivative of \( f(x) = 3x^2 + 4x \): \( F(x) = x^3 + 2x^2 \)
Then: \(\displaystyle \int_2^5 (3x^2 + 4x)\,dx = F(5) – F(2) = (125 + 50) – (8 + 8) = 175 – 16 = 159 \)
Accumulation Functions with Variable Limits
If the upper limit of the accumulation function is a function of \( x \), say \( g(x) \), then by the Chain Rule:
\( \dfrac{d}{dx} \left(\displaystyle \int_a^{g(x)} f(t)\,dt \right) = f(g(x)) \cdot g'(x) \)
This is a direct application of FTC Part 1 plus the Chain Rule when the upper limit is not just \( x \).
Example:
Let \( G(x) =\displaystyle \int_0^{x^2} \sin(t^2)\,dt \). Find \( G'(x) \).
▶️Answer/Explanation
Use Chain Rule: \( G'(x) = \sin((x^2)^2) \cdot \dfrac{d}{dx}(x^2) = \sin(x^4) \cdot 2x \)
Reversed Limits in Accumulation Functions
If the accumulation function is written as \(\displaystyle \int_x^a f(t)\,dt \), you can switch the limits by inserting a negative sign:
\(\displaystyle \int_x^a f(t)\,dt = -\int_a^x f(t)\,dt \)
This allows you to bring the integral back to a form where the upper limit is \( x \) and apply FTC directly.
Example:
Let \( H(x) =\displaystyle \int_x^1 \sqrt{1 + t^2}\,dt \). Find \( H'(x) \).
▶️Answer/Explanation
Rewrite: \( H(x) = -\int_1^x \sqrt{1 + t^2}\,dt \)
Then: \( H'(x) = -\sqrt{1 + x^2} \)
Example:
The figure shows the region \( A \), which is bounded by the \( x \)- and \( y \)-axes, the graph of the function \( f(x) = \dfrac{1 – \cos x}{x} \) for \( x > 0 \), and the vertical line \( x = b \).
If \( b \) increases at a rate of \( \dfrac{\pi}{2} \) units per second, how fast is the area of region \( A \) increasing when \( b = \dfrac{\pi}{3} \)?
▶️Answer/Explanation
The area \( A \) is given by the definite integral:
\( A =\displaystyle \int_0^b \dfrac{1 – \cos x}{x} \, dx \)
We are asked to find \( \dfrac{dA}{dt} \) when \( b = \dfrac{\pi}{3} \), given \( \dfrac{db}{dt} = \dfrac{\pi}{2} \).
By the Fundamental Theorem of Calculus and the Chain Rule:
\( \dfrac{dA}{dt} = \dfrac{1 – \cos b}{b} \cdot \dfrac{db}{dt} \)
Substitute \( b = \dfrac{\pi}{3} \) and \( \dfrac{db}{dt} = \dfrac{\pi}{2} \):
\( \dfrac{dA}{dt} = \dfrac{1 – \cos \left( \dfrac{\pi}{3} \right)}{\dfrac{\pi}{3}} \cdot \dfrac{\pi}{2} \)
Since \( \cos \left( \dfrac{\pi}{3} \right) = \dfrac{1}{2} \), this becomes:
\( \dfrac{dA}{dt} = \dfrac{1 – \dfrac{1}{2}}{\dfrac{\pi}{3}} \cdot \dfrac{\pi}{2} = \dfrac{1}{2} \cdot \dfrac{3}{\pi} \cdot \dfrac{\pi}{2} \)
Simplify:
\( \dfrac{dA}{dt} = \dfrac{1}{2} \cdot \dfrac{3}{\pi} \cdot \dfrac{\pi}{2} = \dfrac{3}{4} \)