Calculus I · Unit 3A · answer key

Unit 3A Practice Exam A Answer Key

Review the worked solutions and modeling rubrics for Unit 3A Practice Exam A. Check methods, units, and interpretation.

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Section overview

Review, practice, exams, and answer keys

What this section is building

Review the worked solutions and modeling rubrics for Unit 3A Practice Exam A. Check methods, units, and interpretation.

Notice

A complete response carries setup, method, computation, bounds or constants, units, interpretation, and an independent verification.

Decide

Attempt the entire problem first, then use one answer at a time to locate the earliest reasoning decision that needs repair.

Avoid

Reading a key before modeling the problem or treating every mismatch as algebra when the first error was conceptual.

Unit 3A Practice Exam A Answer Key

Finish an honest attempt first. Then compare one numbered response at a time, locate the first line where your reasoning diverged, and retry without the key open.

Answer 1

Problem 1

\int\left(4x^3-\frac2x+\sec^2x\right)dx =x^4-2\ln|x|+\tan x+C.

Answer 2

Problem 2

Here $\Delta x=\frac12$ and the midpoints are $\frac14,\frac34,\frac54,\frac74$ . Thus

M_4=\frac12\left[\left(\frac1{16}+1\right)+\left(\frac9{16}+1\right)+\left(\frac{25}{16}+1\right)+\left(\frac{49}{16}+1\right)\right] =\frac{37}{8}.

Answer 3

Problem 3

An indefinite integral is a family of antiderivatives and therefore includes $+C$ . A definite integral is a number defined by a limit of sums; it has fixed bounds and no arbitrary $+C$ in the final value.

Answer 4

Problem 4

By the Fundamental Theorem and chain rule,

G'(x)=e^{\sin^2x}\cos x.

Answer 5

Problem 5

Let $u=1+x^2$ , so $du=2x\,dx$ , and change the bounds from $x=0,1$ to $u=1,2$ :

\int_1^2u^4du=\left[\frac{u^5}{5}\right]_1^2=\frac{31}{5}.

Answer 6

Problem 6

With $u=x$ , $dv=e^{2x}dx$ , and $v=\frac12e^{2x}$ ,

\int xe^{2x}dx=\frac{x e^{2x}}2-\frac{e^{2x}}4+C.

Answer 7

Problem 7

Since

\frac{3x+5}{(x+1)(x+2)}=\frac2{x+1}+\frac1{x+2},

we get

2\ln|x+1|+\ln|x+2|+C.

Answer 8

Problem 8

\int_1^\infty\frac1{x^2}dx =\lim_{b\to\infty}\left[-\frac1x\right]_1^b =\lim_{b\to\infty}\left(1-\frac1b\right)=1.

The integral converges.

Answer 9

Problem 9

The integral is the net change in the amount of water from time 0 to 5 minutes, measured in liters. A negative value means the tank lost more water than it gained over the interval.

Answer 10

Problem 10

Midpoint and trapezoidal rules use constant or linear local approximations; Simpson's Rule uses quadratic pieces and is often more accurate for smooth functions. Every estimate must still be checked against sign, rough area, monotonicity, and a reasonable magnitude.

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Unit 3A Practice Exam A Answer Key

Finish an honest attempt first. Then compare one numbered response at a time, locate the first line where your reasoning diverged, and retry without the key open.

Answer 1

Problem 1

\int\left(4x^3-\frac2x+\sec^2x\right)dx =x^4-2\ln|x|+\tan x+C.

Answer 2

Problem 2

Here $\Delta x=\frac12$ and the midpoints are $\frac14,\frac34,\frac54,\frac74$ . Thus

M_4=\frac12\left[\left(\frac1{16}+1\right)+\left(\frac9{16}+1\right)+\left(\frac{25}{16}+1\right)+\left(\frac{49}{16}+1\right)\right] =\frac{37}{8}.

Answer 3

Problem 3

Answer 4

Problem 4

By the Fundamental Theorem and chain rule,

G'(x)=e^{\sin^2x}\cos x.

Answer 5

Problem 5

Let $u=1+x^2$ , so $du=2x\,dx$ , and change the bounds from $x=0,1$ to $u=1,2$ :

\int_1^2u^4du=\left[\frac{u^5}{5}\right]_1^2=\frac{31}{5}.

Answer 6

Problem 6

With $u=x$ , $dv=e^{2x}dx$ , and $v=\frac12e^{2x}$ ,

\int xe^{2x}dx=\frac{x e^{2x}}2-\frac{e^{2x}}4+C.

Answer 7

Problem 7

Since

\frac{3x+5}{(x+1)(x+2)}=\frac2{x+1}+\frac1{x+2},

we get

2\ln|x+1|+\ln|x+2|+C.

Answer 8

Problem 8

\int_1^\infty\frac1{x^2}dx =\lim_{b\to\infty}\left[-\frac1x\right]_1^b =\lim_{b\to\infty}\left(1-\frac1b\right)=1.

The integral converges.

Answer 9

Problem 9

The integral is the net change in the amount of water from time 0 to 5 minutes, measured in liters. A negative value means the tank lost more water than it gained over the interval.

Answer 10

Problem 10

Source & rights

Original instruction with traceable references.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.

Unit 3A Practice Exam A Answer Key

Related resources

Original instruction with traceable references.

Unit 3A Practice Exam A Answer Key

Related resources

Original instruction with traceable references.