5.8.1.1: Exercises for Section 6.1
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- 186529
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)For exercises 1 - 2, determine the area of the region between the two curves in the given figure by integrating over the \(x\)-axis.
1) \(y=x^2−3\) and \(y=1\)

- Answer
- \(\dfrac{32}{3} \, \text{units}^2\)
2) \(y=x^2\) and \(y=3x+4\)

For exercise 3, split the region between the two curves into two smaller regions, then determine the area by integrating over the \(x\)-axis. Note that you will have two integrals to solve.
3) \(y=x^3\) and \( y=x^2+x\)

- Answer
- \(\dfrac{13}{12}\, \text{units}^2\)
For exercises 4-5, determine the area of the region between the two curves by integrating over the \(y\)-axis.
4) \(x=y^2\) and \(x=9\)

- Answer
- \(36 \, \text{units}^2\)
5) \(y=x\) and \( x=y^2\)

For exercises 6 - 11, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the \(x\)-axis.
6) \(y=x^2\) and \(y=−x^2+18x\)
- Answer
-

243 square units
7) \(y=\dfrac{1}{x}, \quad y=\dfrac{1}{x^2}\), and \(x=3\)
8) \(y=e^x,\quad y=e^{2x−1}\), and \(x=0\)
9) \(y=e^x, \quad y=e^{−x}, \quad x=−1\) and \(x=1\)
- Answer
-

\(\dfrac{2(e−1)^2}{e}\, \text{units}^2\)
10) \( y=e, \quad y=e^x,\) and \(y=e^{−x}\)
11) \(y=|x|\) and \(y=x^2\)
- Answer
-

\(\dfrac{1}{3}\, \text{units}^2\)
For exercises 12 - 15, graph the equations and shade the area of the region between the curves. If necessary, break the region into sub-regions to determine its entire area.
12) \(y=12−x,\quad y=\sqrt{x},\) and \(y=1\)
- Answer
-

\(\dfrac{34}{3}\, \text{units}^2\)
13) \(y=x^3\) and \(y=x^2−2x\) over \(x \in [−1,1]\)
- Answer
-

\(\dfrac{5}{2}\, \text{units}^2\)
14) \(y=x^2+9\) and \( y=10+2x\) over \(x \in [−1,3]\)
15) \(y=x^3+3x\) and \(y=4x\)
- Answer
-

\(\dfrac{1}{2}\, \text{units}^2\)
For exercises 16 - 20, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the \(y\)-axis.
16) \(x=y^3\) and \( x = 3y−2\)
17) \(x=2y\) and \( x=y^3−y\)
- Answer
-

\(\dfrac{9}{2}\, \text{units}^2\)
18) \(x=−3+y^2\) and \( x=y−y^2\)
19) \(y^2=x\) and \(x=y+2\)
- Answer
-

\(\dfrac{9}{2}\, \text{units}^2\)
20) \(x=|y|\) and \(2x=−y^2+2\)
For exercises 21 - 29, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the \(x\)-axis or \(y\)-axis, whichever seems more convenient.
21) \(x=y^4\) and \(x=y^5\)
22) \(y=xe^x,\quad y=e^x,\quad x=0\), and \(x=1\).
- Answer
-

\((e−2)\, \text{units}^2\)
23) \(y=x^6\) and \(y=x^4\)
24) \(x=y^3+2y^2+1\) and \(x=−y^2+1\)
- Answer
-

\(\dfrac{27}{4}\, \text{units}^2\)
25) \( y=|x|\) and \( y=x^2−1\)
26) \(y=4−3x\) and \(y=\dfrac{1}{x}\)
- Answer
-

\(\left(\dfrac{4}{3}−\ln(3)\right)\, \text{units}^2\)
27) \(y=x^2−3x+2\) and \( y=x^3−2x^2−x+2\)
- Answer

\(\dfrac{1}{2}\) square units
28) \(y+y^3=x\) and \(2y=x\)
- Answer
-

\(\dfrac{1}{2}\) square units
29) \( y=\sqrt{1−x^2}\) and \(y=x^2−1\)
For exercises 30 - 37, find the exact area of the region bounded by the given equations if possible. If you are unable to determine the intersection points analytically, use a calculator to approximate the intersection points with three decimal places and determine the approximate area of the region.
30) [T] \(x=e^y\) and \(y=x−2\)
31) [T] \(y=x^2\) and \(y=\sqrt{1−x^2}\)
- Answer
- \(1.067\) square units
32) [T] \(y=3x^2+8x+9\) and \(3y=x+24\)
33) [T] \(x=\sqrt{4−y^2}\) and \( y^2=1+x^2\)
- Answer
- \(0.852\) square units
34) [T] \(x^2=y^3\) and \(x=3y\)
35) [T] \(y=\sqrt{1−x^2}\) and \(y^2=x^2\)
36) [T] \(y=\sqrt{1−x^2}\) and \(y=x^2+2x+1\)
- Answer
- \(\dfrac{3π−4}{12}\) square units
37) [T] \(x=4−y^2\) and \( x=1+3y+y^2\)
38) The largest triangle with a base on the \(x\)-axis that fits inside the upper half of the unit circle \(y^2+x^2=1\) is given by \( y=1+x\) and \( y=1−x\). See the following figure. What is the area inside the semicircle but outside the triangle?

39) A factory selling cell phones has a marginal cost function \(C(x)=0.01x^2−3x+229\), where \(x\) represents the number of cell phones, and a marginal revenue function given by \(R(x)=429−2x.\) Find the area between the graphs of these curves and \(x=0.\) What does this area represent?
- Answer
- $33,333.33 total profit for 200 cell phones sold
40) An amusement park has a marginal cost function \(C(x)=1000e−x+5\), where \(x\) represents the number of tickets sold, and a marginal revenue function given by \(R(x)=60−0.1x\). Find the total profit generated when selling \(550\) tickets. Use a calculator to determine intersection points, if necessary, to two decimal places.
For exercises 41 - 43, find the area between the curves by integrating with respect to \(x\) and then with respect to \(y\). Is one method easier than the other? Do you obtain the same answer?
41) \(y=x^2+2x+1\) and \(y=−x^2−3x+4\)
- Answer
- \(\dfrac{343}{24}\) square units
42) \(y=x^4\) and \(x=y^5\)
43) \(x=y^2−2\) and \(x=2y\)
- Answer
- \(4\sqrt{3}\) square units
For exercises 44 - 45, solve using calculus, then check your answer with geometry.
44) Determine the equations for the sides of the square that touches the unit circle on all four sides, as seen in the following figure. Find the area between the perimeter of this square and the unit circle. Is there another way to solve this without using calculus?

45) Find the area between the perimeter of the unit circle and the triangle created from \(y=2x+1,\,y=1−2x\) and \(y=−\dfrac{3}{5}\), as seen in the following figure. Is there a way to solve this without using calculus?

- Answer
- \( \left(π−\dfrac{32}{25}\right)\) square units
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


