AP Calculus BC 6.7 The Fundamental Theorem of Calculus and Definite Integrals Study Notes - New Syllabus
AP Calculus BC 6.7 The Fundamental Theorem of Calculus and Definite Integrals Study Notes- New syllabus
AP Calculus BC 6.7 The Fundamental Theorem of Calculus and Definite Integrals Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.
Key Concepts:
- The Fundamental Theorem of Calculus and Definite Integrals
The Fundamental Theorem of Calculus and Definite Integrals
The Fundamental Theorem of Calculus and Definite Integrals
The Fundamental Theorem of Calculus (FTC) bridges two central concepts in calculus: the definite integral and the antiderivative. It provides an efficient way to evaluate definite integrals without resorting to Riemann sums or area formulas.
The theorem consists of two key parts:
1. First Part: Derivative of the Accumulation Function
If a function \( f \) is continuous on a closed interval \([a, b]\), and if we define a new function by:
\( F(x) = \displaystyle \int_a^x f(t)\,dt \),
then the function \( F(x) \) is differentiable on \((a, b)\), and its derivative is simply:
\( F'(x) = f(x) \)
This tells us that taking the derivative of the accumulation of \( f \) from \( a \) to \( x \) gives us back the original function \( f(x) \).
2. Second Part: Evaluation of Definite Integrals Using Antiderivatives
If \( f \) is continuous on \([a, b]\), and \( F \) is any antiderivative of \( f \) (i.e., \( F'(x) = f(x) \)), then:
\( \displaystyle \int_a^b f(x)\,dx = F(b) – F(a) \)
This result allows us to evaluate the definite integral by simply finding an antiderivative of \( f(x) \) and applying the limits of integration.
Conclusion: The FTC shows that differentiation and integration are inverse processes. It turns the process of finding areas under curves (definite integrals) into a simpler algebraic calculation using antiderivatives.
Example :
Evaluate the definite integral \( \displaystyle \int_1^4 (3x^2 + 2x)\,dx \).
▶️Answer/Explanation
Step 1: Find the antiderivative of \( f(x) = 3x^2 + 2x \):
\( F(x) = x^3 + x^2 \)
Step 2: Apply the Fundamental Theorem:
\( \displaystyle \int_1^4 (3x^2 + 2x)\,dx = F(4) – F(1) \)
\( = (4^3 + 4^2) – (1^3 + 1^2) = (64 + 16) – (1 + 1) = 80 – 2 = 78 \)
Example:
Evaluate \( \displaystyle \int_0^{\dfrac{\pi}{2}} \cos(x)\,dx \).
▶️Answer/Explanation
Antiderivative of \( \cos(x) \) is \( \sin(x) \)
\( \displaystyle \int_0^{\dfrac{\pi}{2}} \cos(x)\,dx = \sin\left(\dfrac{\pi}{2}\right) – \sin(0) = 1 – 0 = 1 \)
Notes:
- The FTC requires the function to be continuous over the interval.
- Discontinuities may require geometric or limit-based interpretations.
- When evaluating definite integrals, always apply upper limit minus lower limit: \( F(b) – F(a) \).
Example:
Let \( H(x) = \displaystyle \int_{1}^{x^2} \sin(t^3)\,dt \). Find \( H'(x) \).
▶️Answer/Explanation
This is a composition of functions, so we apply the chain rule along with the FTC:
\( H'(x) = \dfrac{d}{dx} \left[ \displaystyle \int_{1}^{x^2} \sin(t^3)\,dt \right] \)
Let the upper limit be \( u = x^2 \), then:
\( H'(x) = \sin((x^2)^3) \cdot \dfrac{d}{dx}(x^2) = \sin(x^6) \cdot 2x \)
So, \( H'(x) = 2x \sin(x^6) \)
Example:
Given \( h(x) = \begin{cases} x – 1 & \text{for } x < 0 \\ \sin x & \text{for } x \ge 0 \end{cases} \), evaluate \( \displaystyle \int_{-1}^{\pi} h(x)\,dx \).
(A) \( \dfrac{3}{2} \)
(B) \( -\dfrac{1}{2} \)
(C) \( -\dfrac{3}{2} \)
(D) \( \dfrac{1}{2} \)
(E) \( -\dfrac{7}{2} \)
▶️Answer/Explanation
We split the integral at 0 due to the piecewise definition:
\( \displaystyle \int_{-1}^{\pi} h(x)\,dx = \displaystyle \int_{-1}^{0} (x – 1)\,dx + \displaystyle \int_{0}^{\pi} \sin x\,dx \)
First part:
\( \displaystyle \int_{-1}^{0} (x – 1)\,dx = \left[\dfrac{1}{2}x^2 – x\right]_{-1}^{0} = \left(0 – 0\right) – \left(\dfrac{1}{2}(-1)^2 – (-1)\right) = -\left(\dfrac{1}{2} + 1\right) = -\dfrac{3}{2} \)
Second part:
\( \displaystyle \int_{0}^{\pi} \sin x\,dx = [-\cos x]_{0}^{\pi} = -\cos(\pi) + \cos(0) = -(-1) + 1 = 2 \)
Total integral:
\( \displaystyle \int_{-1}^{\pi} h(x)\,dx = -\dfrac{3}{2} + 2 = \dfrac{1}{2} \)
Example:
A cubic polynomial is defined as \( f(x) = \dfrac{2}{3}x^3 + ax^2 + bx + c \), where \( a, b, c \) are constants. It has:
- A local minimum at \( x = -2 \)
- A point of inflection at \( x = -5 \)
- \( \displaystyle \int_0^1 f(x)\,dx = \dfrac{15}{2} \)
Find the value of \( c \).
▶️Answer/Explanation
Step 1: Use the critical point \( x = -2 \)
A local minimum implies \( f'(-2) = 0 \). Compute the derivative:
\( f(x) = \dfrac{2}{3}x^3 + ax^2 + bx + c \)
\( f'(x) = 2x^2 + 2ax + b \)
Set \( f'(-2) = 0 \):
\( 2(-2)^2 + 2a(-2) + b = 0 \Rightarrow 8 – 4a + b = 0 \quad \text{(1)} \)
Step 2: Use inflection point at \( x = -5 \)
Point of inflection means \( f”(-5) = 0 \). Compute second derivative:
\( f”(x) = 4x + 2a \)
Set \( f”(-5) = 0 \):
\( 4(-5) + 2a = 0 \Rightarrow -20 + 2a = 0 \Rightarrow a = 10 \quad \text{(2)} \)
Step 3: Substitute \( a = 10 \) into (1):
\( 8 – 4(10) + b = 0 \Rightarrow 8 – 40 + b = 0 \Rightarrow b = 32 \quad \text{(3)} \)
Step 4: Use definite integral condition:
\( \displaystyle \int_0^1 f(x)\,dx = \displaystyle \int_0^1 \left( \dfrac{2}{3}x^3 + 10x^2 + 32x + c \right) dx = \dfrac{15}{2} \)
Integrate term by term:
\( \displaystyle \int_0^1 f(x)\,dx = \left[ \dfrac{1}{6}x^4 + \dfrac{10}{3}x^3 + 16x^2 + cx \right]_0^1 \)
Evaluate at 1:
\( \dfrac{1}{6} + \dfrac{10}{3} + 16 + c = \dfrac{15}{2} \)
Convert everything to sixths:
\( \dfrac{1}{6} + \dfrac{20}{6} + \dfrac{96}{6} + c = \dfrac{15}{2} \Rightarrow \dfrac{117}{6} + c = \dfrac{15}{2} \)
Convert \( \dfrac{15}{2} = \dfrac{45}{6} \):
\( \dfrac{117}{6} + c = \dfrac{45}{6} \Rightarrow c = \dfrac{45 – 117}{6} = \dfrac{-72}{6} = -12 \)
Answer: \( c = -12 \)