2.1E: Exercises for Section 6.1

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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 $y=1$

$This figure is has two graphs. They are the functions f(x) = x^2-3and g(x)=1. In between these graphs is a shaded region, bounded above by g(x) and below by f(x). The shaded area is between x=-2 and x=2.$

Answer: $\dfrac{32}{3} \, \text{units}^2$

2) $y=x^2$ and $y=3x+4$

$This figure is has two graphs. They are the functions f(x) = x^2 and g(x)= 3x+4. In between these graphs is a shaded region, bounded above by g(x) and below by g(x).$

For exercises 3 - 4, 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$

$This figure is has two graphs. They are the functions f(x) = x^3 and g(x)= x^2+x. These graphs intersect twice. The regions between the intersections are shaded. The first region is bounded above by f(x) and below by g(x). The second region is bounded above by g(x) and below by f(x).$

Answer: $\dfrac{13}{12}\, \text{units}^2$

4) $y=\cos θ$ and $ y=0.5$, for $ 0≤θ≤π$

$This figure is has two graphs. They are the functions f(theta) = cos(theta) and g(x)= 0.5. These graphs intersect twice. The regions between the intersections are shaded. The first region is bounded above by f(x) and below by g(x). The second region is bounded above by g(x) and below by f(x).$

For exercises 5-6, determine the area of the region between the two curves by integrating over the $y$-axis.

5) $x=y^2$ and $x=9$

$This figure is has two graphs. They are the equations x=y^2 and x=9. The region between the graphs is shaded. It is horizontal, between the y-axis and the line x=9.$

Answer: $36 \, \text{units}^2$

6) $y=x$ and $ x=y^2$

$This figure is has two graphs. They are the equations y=x and x=y^2. The region between the graphs is shaded, bounded above by x=y^2 and below by y=x.$

For exercises 7 - 13, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the $x$-axis.

7) $y=x^2$ and $y=−x^2+18x$

Answer

$This figure is has two graphs. They are the functions f(x)=x^2 and g(x)=-x^2+18x. The region between the graphs is shaded, bounded above by g(x) and below by f(x). It is in the first quadrant.$

243 square units

8) $y=\dfrac{1}{x}, \quad y=\dfrac{1}{x^2}$, and $x=3$

9) $y=\cos x$ and $y=\cos^2x$ on $x \in [−π,π]$

Answer

$This figure is has two graphs. They are the functions y=cos(x) and y=cos^2(x). The graphs are periodic and resemble waves. There are four regions created by intersections of the curves. The areas are shaded.$

4 square units

10) $y=e^x,\quad y=e^{2x−1}$, and $x=0$

11) $y=e^x, \quad y=e^{−x}, \quad x=−1$ and $x=1$

Answer

$This figure is has two graphs. They are the functions f(x)=e^x and g(x)=e^-x. There are two shaded regions. In the second quadrant the region is bounded by x=-1, g(x) above and f(x) below. The second region is in the first quadrant and is bounded by f(x) above, g(x) below, and x=1.$

$\dfrac{2(e−1)^2}{e}\, \text{units}^2$

12) $ y=e, \quad y=e^x,$ and $y=e^{−x}$

13) $y=|x|$ and $y=x^2$

Answer

$This figure is has two graphs. They are the functions f(x)=x^2 and g(x)=absolute value of x. There are two shaded regions. The first region is in the second quadrant and is between g(x) above and f(x) below. The second region is in the first quadrant and is bounded above by g(x) and below by f(x).$

$\dfrac{1}{3}\, \text{units}^2$

For exercises 14 - 19, 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.

14) $y=\sin(πx),\quad y=2x,$ and $x>0$

15) $y=12−x,\quad y=\sqrt{x},$ and $y=1$

Answer

$This figure is has three graphs. They are the functions f(x)=squareroot of x, y=12-x, and y=1. The region between the graphs is shaded, bounded above and to the left by f(x), above and to the right by the line y=12-x, and below by the line y=1. It is in the first quadrant.$

$\dfrac{34}{3}\, \text{units}^2$

16) $y=\sin x$ and $y=\cos x$ over $x \in [−π,π]$

17) $y=x^3$ and $y=x^2−2x$ over $x \in [−1,1]$

Answer

$This figure is has two graphs. They are the functions f(x)=x^3 and g(x)=x^2-2x. There are two shaded regions between the graphs. The first region is bounded to the left by the line x=-2, above by g(x) and below by f(x). The second region is bounded above by f(x), below by g(x) and to the right by the line x=2.$

$\dfrac{5}{2}\, \text{units}^2$

18) $y=x^2+9$ and $ y=10+2x$ over $x \in [−1,3]$

19) $y=x^3+3x$ and $y=4x$

Answer

$This figure is has two graphs. They are the functions f(x)=x^3+3x and g(x)=4x. There are two shaded regions between the graphs. The first region is bounded above by f(x) and below by g(x). The second region is bounded above by g(x), below by f(x).$

$\dfrac{1}{2}\, \text{units}^2$

For exercises 20 -25, graph the equations and shade the area of the region between the curves. Determine its area by integrating over the $y$-axis.

20) $x=y^3$ and $ x = 3y−2$

21) $x=2y$ and $ x=y^3−y$

Answer

$This figure is has two graphs. They are the equations x=2y and x=y^3-y. The graphs intersect in the third quadrant and again in the first quadrant forming two closed regions in between them.$

$\dfrac{9}{2}\, \text{units}^2$

22) $x=−3+y^2$ and $ x=y−y^2$

23) $y^2=x$ and $x=y+2$

Answer

$This figure is has two graphs. They are the equations x=y+2 and y^2=x. The graphs intersect, forming a region in between them$

$\dfrac{9}{2}\, \text{units}^2$

24) $x=|y|$ and $2x=−y^2+2$

25) $x=\sin y,\quad x=\cos(2y),\quad y=π/2$, and $ y=−π/2$

Answer

$This figure is has two graphs. They are the equations x=cos(y) and x=sin(y). The graphs intersect, forming two regions bounded above by the line y=pi/2 and below by the line y=-pi/2.$

$\dfrac{3\sqrt{3}}{2}\, \text{units}^2$

For exercises 26 - 37, 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.

26) $x=y^4$ and $x=y^5$

27) $y=xe^x,\quad y=e^x,\quad x=0$, and $x=1$.

Answer

$This figure is has two graphs. They are the equations y=xe^x and y=e^x. The graphs intersect, forming a region in between them in the first quadrant.$

$(e−2)\, \text{units}^2$

28) $y=x^6$ and $y=x^4$

29) $x=y^3+2y^2+1$ and $x=−y^2+1$

Answer

$Two curves plotted against X and Y axes. One curve is a sideways parabola opening to the left and labeled 'X equals negative Y squared plus 1.' The other curve looks like a horizontally stretched letter S and is labeled 'X equals Y cubed plus 2 Y squared plus 1.' The apex of the parabola touches the upper curve of the S near X equals 1 and Y equals 0. The lower curve of the S also crosses the parabola near X equals negative 8 and Y equals negative 3.$

$\dfrac{27}{4}\, \text{units}^2$

30) $ y=|x|$ and $ y=x^2−1$

31) $y=4−3x$ and $y=\dfrac{1}{x}$

Answer

$This figure is has two graphs. They are the equations y=4-3x and y=1/x. The graphs intersect, having region between them shaded. The region is in the first quadrant.$

$\left(\dfrac{4}{3}−\ln(3)\right)\, \text{units}^2$

32) $y=\sin x,\quad x=−π/6,\quad x=π/6,$ and $y=\cos^3 x$

33) $y=x^2−3x+2$ and $ y=x^3−2x^2−x+2$

Answer: $This figure is has two graphs. They are the equations y=x^2-3x+2 and y=x^3-2x^2-x+2. The graphs intersect, having region between them shaded.$
$\dfrac{1}{2}$ square units

34) $y=2\cos^3(3x),\quad y=−1,\quad x=\dfrac{π}{4},$ and $ x=−\dfrac{π}{4}$

35) $y+y^3=x$ and $2y=x$

Answer

$This figure is has two graphs. They are the equations 2y=x and y+y^3=x. The graphs intersect, forming two regions. The regions are shaded.$

$\dfrac{1}{2}$ square units

36) $ y=\sqrt{1−x^2}$ and $y=x^2−1$

37) $y=\cos^{−1}x,\quad y=\sin^{−1}x,\quad x=−1,$ and $ x=1$

Answer

$This figure is has two graphs. They are the equations y=arccos(x) and y=arcsin (x). The graphs intersect, forming two regions. The first region is bounded to the left by x=-1. The second region is bounded to the right by x=1. Both regions are shaded.$

$−2(\sqrt{2}−π)$ square units

For exercises 38 - 47, 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.

38) [T] $x=e^y$ and $y=x−2$

39) [T] $y=x^2$ and $y=\sqrt{1−x^2}$

Answer: $1.067$ square units

40) [T] $y=3x^2+8x+9$ and $3y=x+24$

41) [T] $x=\sqrt{4−y^2}$ and $ y^2=1+x^2$

Answer: $0.852$ square units

42) [T] $x^2=y^3$ and $x=3y$

43) [T] $y=\sin^3x+2,\quad y=\tan x,\quad x=−1.5,$ and $x=1.5$

Answer: $7.523$ square units

44) [T] $y=\sqrt{1−x^2}$ and $y^2=x^2$

45) [T] $y=\sqrt{1−x^2}$ and $y=x^2+2x+1$

Answer: $\dfrac{3π−4}{12}$ square units

46) [T] $x=4−y^2$ and $ x=1+3y+y^2$

47) [T] $y=\cos x,\quad y=e^x,\quad x=−π,\quad$ and$\quad x=0$

Answer: $1.429$ square units

48) 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?

$This figure is has the graph of a circle with center at the origin and radius of 1. There is a triangle inscribed with base on the x-axis from -1 to 1 and the third corner at the point y=1.$

49) 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

50) 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.

51) The tortoise versus the hare: The speed of the hare is given by the sinusoidal function $H(t)=1−\cos((πt)/2)$ whereas the speed of the tortoise is $T(t)=(1/2)\tan^{−1}(t/4)$, where $t$ is time measured in hours and the speed is measured in miles per hour. Find the area between the curves from time $t=0$ to the first time after one hour when the tortoise and hare are traveling at the same speed. What does it represent? Use a calculator to determine the intersection points, if necessary, accurate to three decimal places.

Answer: $3.263$ mi represents how far ahead the hare is from the tortoise

52) The tortoise versus the hare: The speed of the hare is given by the sinusoidal function $H(t)=(1/2)−(1/2)\cos(2πt)$ whereas the speed of the tortoise is $T(t)=\sqrt{t}$, where $t$ is time measured in hours and speed is measured in kilometers per hour. If the race is over in 1 hour, who won the race and by how much? Use a calculator to determine the intersection points, if necessary, accurate to three decimal places.

For exercises 53 - 55, 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?

53) $y=x^2+2x+1$ and $y=−x^2−3x+4$

Answer: $\dfrac{343}{24}$ square units

54) $y=x^4$ and $x=y^5$

55) $x=y^2−2$ and $x=2y$

Answer: $4\sqrt{3}$ square units

For exercises 56 - 57, solve using calculus, then check your answer with geometry.

56) 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?

$This figure is the graph of a circle centered at the origin with radius of 1. There is a circumscribed square around the circle.$

57) 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?

$This figure is the graph of a circle centered at the origin with radius of 1. There are three lines intersecting the circle. The lines intersect the circle at three points to form a triangle within the circle.$

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.

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