MMA Math 2024-25 AP Calculus BC Unit 6 Practice Test PDF

# MMA Math 2024-25 | AP Calculus BC - Unit 6 Practice Test ## Name: ANNE SERKY ## Block: ∞ # Practice Test - Unit 6: Integration and the Fundamental Theorem of Calculus **Instructions:** Answer every question as best you can and show all work. Check the solutions when posted and come prepared...

# MMA Math 2024-25 | AP Calculus BC - Unit 6 Practice Test ## Name: ANNE SERKY ## Block: ∞ # Practice Test - Unit 6: Integration and the Fundamental Theorem of Calculus **Instructions:** Answer every question as best you can and show all work. Check the solutions when posted and come prepared with questions on the review day. All questions on this practice test can be solved without a calculator. ## 1. Determine whether each statement is true or false. **I.** The function *g(x) = ∫_a^x *f(t) dt is increasing for all values of x for which the graph of y= f(x) is above the x-axis. **True/False** *g(x) is increasing whenever *g'(x) > 0 and *g'(x) = f(x) by the FTC Part 1* **II.** The integral ∫_a^b *v(t) dt, where *v(t) is the velocity of a particle moving along a straight line, gives the particle's change in position on [a, b]. **True/False** *By definition of velocity and resulting from the FTC Part 2.* **III.** If a function *f* is continuous and positive on the interval [a, b], it's guaranteed that there exists a number c ∈ [a, b] such that the area of the rectangle *f(c)(b-a)* equals ∫_a^b *f(x) dx*. **True/False** *MVT for Integrals* ## 2. Find all values of x for which the given function g has a local maximum. *g(x) = ∫₁^x *√(t² - 1) dt* *g'(x) = ∫₁^x *√(t²- 1) dt* *g'(x) = √(x² - 1) * 2x* **Set *g'(x) = 0* *x = -1, x = 0, x = 1* *g'(-2) = √((-2)² - 1) * 2 (-2) = -2√3* *g'(1/2) = √((1/2)² -1) * 2 (1/2) = -√3/2* *g'(2) = √(2² - 1) * 2 (2) = 2√3* **C.P.'s** *g'(-2) < 0, *g'(1/2) < 0, *g'(2) > 0* *g' - - + + + + + + + + + + + + + + + -2 0 1 *g' has a local maximum at x = 0. Since g' changes from positive to negative* ## 3. The acceleration of a particle moving along a horizontal coordinate is given by *a(t) = cos(2t), t ≥ 0*. At t = 0, the velocity of the particle is 0 and its position is 0. What is the position of the particle at t = π/4, and is the particle moving toward or away from the origin? *a(t) = cos(2t)* *v(0) = 0* *s(0) = 0* *v(t) = ∫cos(2t) dt* *v(t) = (1/2)sin(2t) + C* *v(0) = 0 → 0 = (1/2)sin(2(0)) + C → C = 0* *v(t) = (1/2)sin(2t)* *v(π/4) = (1/2)sin(2(π/4)) = (1/2)* *s(t) = ∫v(t) dt* *s(t) = ∫(1/2)sin(2t) dt* *s(t) = (-1/4)cos(2t) + C* *s(0) = 0 → 0 = (-1/4)cos(2(0)) + C → C = (1/4)* *s(t) = (-1/4)cos(2t) + (1/4)* *s(π/4) = (-1/4)cos(2(π/4)) + (1/4) = (1/2)* *The particle is moving away at t = π/4 because both position and velocity are positive.* ## 4. Consider the region between the graph of *f(x) = 2e^-x² and the x-axis for -1 < x < 1, as shown in the figure below. *a) Find, but do not solve, a definite integral that gives the area of the region.* *A = ∫_-1¹ 2e^-x² dx* *b) Express the definite integral found in part (a) as the limit of a Riemann Sum.* ∫_-1¹ 2e^-x² dx = lim_n→∞ Σ_k=1ⁿ 2e^{-(-1 + k/n)²} (2/n) *c) If A denotes the area of the region, express the average value of f over -1 < x < 1 in terms of A.* f_AVG = 1/(b - a) ∫_a^bf(x)dx f_AVG = (1/2)A ## 5. Evaluate. *a) ∫₀^π/3 cos(x/2) dx = 2sin(x/2)|₀^π/3 = 2sin(π/6) - 2sin(0) = 2sin(π/6)-0 = (2)(1/2) = 1* *b) ∫₂¹⁰ 6/(4x - 4) dx = ∫₂¹⁰ 6/(4(x - 1)) = (3/2)∫₂¹⁰ 1/(x - 1) dx = (3/2)[ln(|x - 1|)]|₂¹⁰ = (3/2)[ln(|10 - 1|) - ln(|2 - 1|)] = (3/2)[ln(9) - ln(1)] = (3/2)ln(9)* *c) ∫₁² e^-2x dx = (-1/2)e^-2x|₁² = (-1/2)(e^-2(2) - e^-2(1)) = (-1/2)(e^-4 - e^-2) = (1/2)(e^-2 - e^-4)* ## 6. Let *f* be a continuous function on [-5, 6] graphed in the figure below. Its graph consists of a semi-circle of radius 3 and two straight line segments. **Graph of f** The function *g* is given by *g(x) = ∫₀^xf(t) dt.* Find the absolute maximum and absolute minimum values of g on [-5, 6], stating the x-values at which they occur. Justify your answers **Need to Test:** - **Endpoints:** x = -5, x = 6 - **Extrema (Where f' changes signs)** x = -3 *g(6) = ∫₀⁶f(t)dt = ∫₀³f(t)dt + ∫₃⁶f(t)dt = (1/4)π(3)² + (1/2)(3)(4) = 9 + 6 = 15 * *g(-5) = ∫₀^-5f(t)dt = -∫_-5⁰f(t)dt = -∫_-5^-3f(t)dt - ∫_-3⁰f(t)dt = -[(-2)(3) + (1/4)π(3)²] = 2 - (9/4)π* *g(-3) = ∫₀^-3f(t)dt = -∫_-3⁰f(t)dt = -(9/4)π* *Abs. Max = 15 at x = 6* *Abs. Min. = -(9/4)π at x = -3* ## 7. The derivative of a differentiable function f is graphed in the figure below. **Graph of y = f'(x)** A = (1/2)h(b₁ + b₂) *a) If f(0) = 8, what is the value of f(8)?* ∫₀⁸ f'(x) dx = f(8) - f(0) f(8) = f(0) + ∫₀⁸ f'(x) dx = 8 + (1/2)(8)(2 + 6) = 8 + 32 = 40 *b) What is the slope of the line tangent to the graph of f at x = 8?* m = f'(8) f'(8) = 6 *The slope is 6.*