Math 152: Worksheet 2

Area Between Two Curves

Learning Problems

These problems should be completed on your own. If you need hints on solving a problem, there are some provided with each problem. Click on the word "hint" to view it and again to hide it. They go in increasing order of helpfulness, with the last hint mostly giving away how to do the problem. Try to work from the earlier hints to the later ones, as this will give you the practice you need to succeed in this class.

Problem 1

Find the area between the curves $y = \sin(x)$ and $y = x+1$ over the interval $[0, \pi]$.

• Hint 1

  Sketch a graph of these two functions. Do these curves cross?

• Hint 2

  Plugging in 0 or 1 tells us that $x+1$ is on top and $\sin{x}$ is on bottom. What should the integral for the area look like?

• Hint 3

  The integral should be $\displaystyle \int_0^\pi (x+1) - \sin{x}\ dx$

• Answer

  $\displaystyle \frac{\pi^2}{2} + \pi - 2$

Problem 2

Find the area of the region bounded by the curves $y = 2x+1$ and $y = 5 + 2x - x^2$.

• Hint 1

  Where do these curves cross? You should end up getting two points. Then which function is on top?

• Hint 2

  The curves cross at $2$ and $-2$, and the parabola is on top over this domain. Use that to set up the integral.

• Hint 3

  The integral should be $\displaystyle \int_{-2}^2 (5 + 2x - x^2) - (2x + 1)\ dx$

• Answer

  $\displaystyle \frac{32}{3}$

Problem 3

Find the area of the region bounded by the curves $y = x^2$, $y=1$, and $y = 6-x$ that contains the point $(2,2)$.

• Hint 1

  Sketch out a picture. Is this region vertically simple? What about horizontally simple?

• Hint 2

  Figure out where these curves cross. You can either set up the problem as a single integral in $y$, or two integrals in $x$.

• Hint 3

  If you want to solve it as an integral in $y$, remember to solve out for $x$ as a function of $y$ in each case.

• Hint 4

  The integral you get should either be $\displaystyle \int_1^2 x^2 - 1\ dx + \int_2^5 (6-x) - 1\ dx$ or $\displaystyle \int_1^4 (6-y) - \sqrt{y}\ dy$

• Answer

  $\displaystyle \frac{35}{6}$

Problem 4

Find the area of the region bounded by the curves $x = y^2 - 2y - 1$ and $x - 2y = 4$.

• Hint 1

  Figure out where these two curves cross. The first function should indicate how you want to carry out this integral.

• Hint 2

  Solving for x in each expression and setting them equal gives that they cross at $y=5$ and $y=-1$.

• Hint 3

  Which function is farther right, and thus more positive? Plug in $0$ to find out.

• Hint 4

  The integral should come out to $\displaystyle \int_{-1}^5 (2y+4) - (y^2 - 2y - 1)\ dy$.

• Answer

  $36$

Problem 5

Find the area bounded between the curves $y=2x$, $y = x^2 + 2x$ and $x+y = 8$.

• Hint 1

  Sketch a picture and see what this region looks like. Where do all of these curves intersect?

• Hint 2

  You should get intersections points of $(0,0)$, $(\frac{10}{3}\frac{20}{3})$ and $(2,8)$.

• Hint 3

  You can set this area up as either two integrals in y or two integrals in x. Either way it will require two integrals, but solving out for $x$ in the one expression is kind of annoying.

• Hint 4

  The integrals should be $\displaystyle \int_0^2 (x^2 + 2x) - 2x\ dx + \int_2^{10/3} (8-x) - x\ dx $ or $\displaystyle \int_0^{20/3} (8-y) - \frac{y}{2}\ dy + \int_{20/3}^8 (8-y) - (\sqrt{y+1} - 1)\ dy$

• Answer

  $\displaystyle \frac{56}{9}$

Submission Problems

Problem 1

Find the area bounded by the curves $y = x^2 - 5$ and $y = 3 - 6x - x^2$.

Problem 2

Find the area of the triangle defined by the equations $y = x$, $y = 5x$ and $y = 12 - 3x$