Practice Problems

Site: Saylor Academy
Course: MA005: Calculus I
Book: Practice Problems
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Date: Monday, July 15, 2024, 10:39 AM

Description

Work through the odd-numbered problems 1-25. Once you have completed the problem set, check your answers.

Table of contents

Practice Problems

In problems 1 – 3, the function f is given by a graph, and \mathrm{A}(\mathrm{x})=\int_{1}^{x} \mathrm{f}(t) \mathrm{dt}.

(a) Graph y=A(x) \text { for } 1 \leq x \leq 5.

(b) Estimate the values of \mathrm{A}(1), \mathrm{A}(2), \mathrm{A}(3), \text { and } \mathrm{A}(4).

(c) Estimate the values of A^{\prime}(1), A^{\prime}(2), A^{\prime}(3), \text { and } A^{\prime}(4).

1. f in Fig. 11.

3. f in Fig. 13


In problems 5 – 7, the function f is given by a formula, and \mathrm{A}(\mathrm{x})=\int_{1}^{x} \mathrm{f}(t) \mathrm{dt}.

(a) Graph y=A(x) and 1 \leq x \leq 5.

(b) Calculate the values of  \mathrm{A}(1), \mathrm{A}(2), \mathrm{A}(3), \text { and } \mathrm{A}(4).

(c) Determine the values of \mathrm{A}^{\prime}(1), \mathrm{A}^{\prime}(2), \mathrm{A}^{\prime}(3) and A^{\prime}(4).

5. \mathrm{f}(t)=2

7.  \mathrm{f}(t)=6-t


In problems 9 – 17, use the Antiderivatives and Definite Integrals Theorem to evaluate the integrals.

9. \begin{array}{ll}
    \int_{0}^{3} 2 x \mathrm{dx}, & \int_{1}^{3} 2 x \mathrm{~d} \mathrm{x}, & \int_{0}^{1} 2 x \mathrm{~d} \mathrm{x}
    \end{array}

11. \int_{1}^{3} 6 x^{2} \mathrm{dx}, \quad \int_{1}^{2} 6 x^{2} \mathrm{dx}, \quad \int_{0}^{3} 6 x^{2} \mathrm{dx}

13. \begin{array}{ll}
    \int_{0}^{3} 4 x^{3} \mathrm{dx}, & \int_{1}^{3} 4 x^{3} \mathrm{dx}, & \int_{0}^{1} 4 x^{3} \mathrm{dx} \\
    \end{array}

15. \begin{aligned}
    &\int_{-3}^{3} 3 x^{2} \mathrm{dx}, \quad \int_{-3}^{0} 3 x^{2} \mathrm{dx}, \quad \int_{0}^{3} 3 x^{2} \mathrm{dx} \\
    &
    \end{aligned}

17. \int_{0}^{2} 3 x^{2} \mathrm{dx}, \quad \int_{1}^{3} 3 x^{2} \mathrm{dx}


19. The velocity of a car after t seconds is 2 t feet per second. (a) How far does the car travel during its first 10 seconds? (b) How many seconds does it take the car to travel half the distance in part (a)?

21. The velocity of a car after t seconds is 4 t^{3} feet per second. (a) How far does the car travel during its first 10 seconds? (b) How many seconds does it take the car to travel half the distance in part (a)?

23. The velocity of a car after t seconds is 75-3 t^{2} feet per second. (a) How many seconds does it take for the car to come to a stop (velocity = 0)? (b) How far does the car travel while coming to a stop? (c) How many seconds does it take the car to travel half the distance in part (b)?

25. An artist you know wants to make a figure consisting of the region between the curve y=x^{2} and the x–axis for 0 \leq x \leq 3.

(a) Where should the artist divide the region with a vertical line (Fig. 15) so each piece has the same area?

(b) Where should the artist divide the region with vertical lines to get 3 pieces with equal areas?


Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-5.5-Areas-Integrals-Antiderivatives.pdf
Creative Commons License This work is licensed under a Creative Commons Attribution 3.0 License.

Answers

1.

(a) See Figure.

(b) \mathrm{A}(1)=0, \mathrm{~A}(2)=1.5, \mathrm{~A}(3)=4, \mathrm{~A}(4)=6.5

(c) \text { (c) } \mathrm{A}^{\prime}(1)=1, \mathrm{~A}^{\prime}(2)=2, \mathrm{~A}^{\prime}(3)=3, \mathrm{~A}^{\prime}(4)=2

3.

(a) see Figure

(b) \mathrm{A}(1)=0, \mathrm{~A}(2)=0.5, \mathrm{~A}(3)=0, \mathrm{~A}(4)=-1

(c) \mathrm{A}^{\prime}(1)=1, \mathrm{~A}^{\prime}(2)=0, \mathrm{~A}^{\prime}(3)=-1, \text { A }^{\prime}(4)=-1

5.

(a) see Figure

(b)\mathrm{A}(1)=0, \mathrm{~A}(2)=2, \mathrm{~A}(3)=4, \mathrm{~A}(4)=6

(c) A^{\prime}(1)=2, A^{\prime}(2)=2, A^{\prime}(3)=2, A^{\prime}(4)=2

7.

(a) see Figure

(b) \mathrm{A}(1)=0, \mathrm{~A}(2)=4.5, \mathrm{~A}(3)=8, \mathrm{~A}(4)=10.5

(c) A^{\prime}(1)=5, A^{\prime}(2)=4, A^{\prime}(3)=3, A^{\prime}(4)=2


9. \left.\mathrm{x}^{2}\right|_{0} ^{3}=9,\left.\mathrm{x}^{2}\right|_{1} ^{3}=8,\left.\mathrm{x}^{2}\right|_{0} ^{1}=1

11. \left.2 \mathrm{x}^{3}\right|_{1} ^{3}=52,\left.2 \mathrm{x}^{3}\right|_{1} ^{2}=14,\left.2 \mathrm{x}^{3}\right|_{0} ^{3}=54

13. \left.\mathrm{x}^{4}\right|_{0} ^{3}=81,\left.\mathrm{x}^{4}\right|_{1} ^{3}=80,\left.\mathrm{x}^{4}\right|_{0} ^{1}=1

15. \left.\mathrm{x}^{3}\right|_{-3} ^{3}=54,\left.\mathrm{x}^{3}\right|_{-3} ^{0}=27,\left.\mathrm{x}^{3}\right|_{0} ^{3}=27

17. \left.x^{3}\right|_{0} ^{2}=8,\left.x^{3}\right|_{1} ^{3}=26


19.

(a) distance = \int_{0}^{10} 2 \mathrm{t} \mathrm{dt}=\left.\mathrm{t}^{2}\right|_{0} ^{10}=100 feet.

(b) Find T so 50=\int_{0}^{\mathrm{T}} 2 \mathrm{t} \mathrm{dt}=\left.\mathrm{t}^{2}\right|_{0} ^{\mathrm{T}}=\mathrm{T}^{2} . \mathrm{T}=\sqrt{50} \approx 7.07 seconds.


21.

(a) distance = \int_{0}^{10} 4 \mathrm{t}^{3} \mathrm{dt}=\left.\mathrm{t}^{4}\right|_{0} ^{10}=10,000 feet.

(b) 5000=\int_{0}^{T} 4 t^{3} \mathrm{dt}=\mathrm{T}^{4} \cdot \mathrm{T}=\sqrt[4]{5000} \approx 8.41 \text { seconds }.


23.

(a) velocity = 75-3 t^{2}=0 when t = 5 seconds. (b) distance = \int_{0}^{5} 75-3 t^{2} \mathrm{dt}=75 \mathrm{t}-\left.\mathrm{t}^{3}\right|_{0} ^{5}=250 feet.

(b) 125=\int_{0}^{T} 75-3 \mathrm{t}^{2} \mathrm{dt}=75 \mathrm{t}-\left.\mathrm{t}^{3}\right|_{0} ^{\mathrm{T}}=75 \mathrm{~T}-\mathrm{T}^{3} \text { so } \mathrm{T}^{3}-75 \mathrm{~T}+125=0 (solve using Newton's method or by examining the graph of y=x^{3}-75 x+125) and \mathrm{T} \approx 1.74 seconds.


25. The total area is \int_{0}^{3} x^{2} d x=\left.\frac{1}{3} x^{3}\right|_{0} ^{3}=9.

(a) Find T so \frac{1}{2} \cdot 9=\frac{9}{2}=\int_{0}^{\mathrm{T}} \mathrm{x}^{2} \mathrm{dx}=\left.\frac{1}{3} \mathrm{x}^{3}\right|_{0} ^{\mathrm{T}}=\frac{1}{3} \mathrm{~T}^{3} \cdot \mathrm{T}=\sqrt[3]{27 / 2} \approx 2.38.

(b) Find T so \frac{1}{3} \cdot 9=3=\int_{0}^{T} \mathrm{x}^{2} \mathrm{dx}=\left.\frac{1}{3} \mathrm{x}^{3}\right|_{0} ^{\mathrm{T}}=\frac{1}{3} \mathrm{~T}^{3} \cdot \mathrm{T}=\sqrt[3]{9} \approx 2.08 .

Then find T so \frac{2}{3} \cdot 9=6=\int_{0}^{\mathrm{T}} \mathrm{x}^{2} \mathrm{dx}=\left.\frac{1}{3} \mathrm{x}^{3}\right|_{0} ^{\mathrm{T}}=\frac{1}{3} \mathrm{~T}^{3} \cdot \mathrm{T}=\sqrt[3]{18} \approx 2.62.