MAT 210 Brief Calculus - Sample Final Exam (Practice Version A) - Solutions

Sep 30, 2026

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MAT 210 Brief Calculus Sample Final Exam (Practice Version A) provides a set of multiple-choice problems covering definite integrals, antiderivatives, properties of reversing integration limits, Riemann sums, critical points when derivatives do not exist, and optimization of revenue from a nonlinear demand model. The solutions section walks through key calculus techniques: setting up area integrals between curves, evaluating definite integrals, applying the sign change of swapped limits, constructing left Riemann sums, and using derivatives to find maxima. It also includes examples involving concavity and monotonicity via first and second derivatives, computing an original profit function from its derivative, and determining instantaneous growth rate for a logarithmic bacterial model.

LanguageEnglish

CategoryExam

FormatPDF

Pages14

Reading Time35 min

AccessFree to download

Multiple Choice

Problems 1-9

MAT 210 Brief Calculus Sample Final Exam (Practice Version A) Solutions | Instructor | XXXXXXX | | --- | --- | | Full Name (Print clearly!) | | | Student ID | | Important Disclaimer for Students: • This is a sample Final exam, not the actual in-class exam. Questions are of similar difficulty but may differ in type, numbers, or context. • Success in the course depends on practicing concepts beyond this sample: lecture notes, Edfinity assignments, group work, quizzes, and homework. • Complete this exam under exam-like conditions: no notes, no internet, no communication, and about 120- 140 minutes to finish. • After finishing, check your answers against the posted solutions and grade yourself honestly. • Repeat for all other versions posted on the MAT 210 website. • If you cannot complete it within the time limit, review your weak areas and try another sample exam. • Round your answers to two decimal places where needed. • Treat all concepts from class materials as fair game for the actual exam. Part I Multiple Choice: Circle the response that best completes the statement or answers the question. Then write the letter of your answer in the table after Problem 10 . 1. (5 points) Identify which of the following definite integrals corresponds to the area bounded by the curves y = 4x + 16 and y = 2x2 + 10 shown below. A. Z−31(4x − 2x2 + 6) dx B. Z−31 􀀂(2x2 + 10) − (4x + 16) 􀀃 dx C. Z04 (4x − 2x2 + 6) dx D. Z−31 􀀂(4x + 16) + (2x2 + 10) 􀀃 dx E. None of the above Figure 1 Solution: The area between two curves is given by: Area = Zxx12 􀀂top curve − bottom curve 􀀃 dx • The curves intersect at x = −1 and x = 3 (solve 4x + 16 = 2x2 + 10) . • For x ∈ [−1 , 3], the line y = 4x + 16 lies above the parabola y = 2x2 + 10. • Therefore, the integral for the area is: Z−31 􀀂(4x + 16) − (2x2 + 10) 􀀃dx = Z−31(−2x2 + 4x + 6) dx Correct answer: A 2. (5 points) Evaluate 4 Z (3x2 − 2x) dx. 1 A. 33 B. 48 C. 30 D. 24 E. None of the above Solution: Step 1: Find the antiderivative of the integrand: Z (3x2 − 2x)dx = x3 − x2 + C Step 2: Evaluate the definite integral from 1 to 4: Z14 (3x2 − 2x) dx = 􀀂x3 − x2 􀀃 41 = (43 − 42 ) − (13 − 12 ) Step 3: Simplify: (64 − 16) − (1 − 1) = 48 − 0 = 48 Correct Answer: B 3. (5 points) Which of the following is true for any continuous function f (x) on [a, b]? A. Za b f (x) dx + Zb a f (x) dx = 0 B. Za b f (x) dx = 0 C. Za b f (x) dx = Zab |f(x)| dx D. Za b f (x) dx = f(b) − f(a) E. None of the above Solution: Property: Reversing the limits of integration changes the sign of the integral: Zb a f (x) dx = − Za b f (x) dx. Therefore, Za b f (x) dx + Zb a f (x) dx = Za b f (x) dx − Za b f (x) dx = 0 . Option A is always true for any continuous function. Correct Answer: A 4. (5 points) For f (x) = x3 on [0 , 2] with ∆x = 0 .5, the Left Riemann sum is given by: A. 0.5(f(0) + f(0 .5) + f(1) + f(1 .5)) B. 0.5(f(0 .5) + f(1) + f(1 .5) + f(2)) C. 0.5(f(0) + f(0 .5) + f(1) + f(1 .5) + f(2)) D. 2(f(0) + f(0 .5) + f(1) + f(1 .5)) E. None of the above Solution: A Left Riemann sum on the interval [0 , 2] with ∆x = 0 .5 uses the left endpoints: 0 , 0.5 , 1 , 1.5 So the correct expression is: 0.5(f(0) + f(0 .5) + f(1) + f(1 .5)) . Correct Answer: A. 5. (5 points) If h(2) exists, what can we conclude if h′(2) does not exist? A. x = 2 cannot be a critical point since the derivative does not exist. B. x = 2 must be a relative maximum or a relative minimum. C. x = 2 is a critical point. D. x = 2 cannot be a relative extrema point. E. Nothing can be concluded without more information about the function. | Solution: A critical point occurs where the derivative is zero or does not exist, provided the function is defined at that point. Since h(2) exists and h′(2) does not exist, x = 2 is a critical point. | | --- | | 6. (5 points) A small business models revenue with a nonlinear demand f

FAQs

How is the area between two curves computed in Problem 1?

Area is calculated as the integral of the top curve minus the bottom curve over the intersection interval where the curves meet.

What method is used to evaluate the definite integral in Problem 2?

Compute an antiderivative of the integrand, then evaluate it at the upper and lower bounds and subtract.

How is instantaneous rate of change found in Problem 9?

Differentiate the given function to obtain the derivative, then evaluate the derivative at the specified time value.

Sep 30, 2026

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