Area Under Curve: Calculate Integral from 0 to 1

[karush]

A region is bounded between the graphs of y = -1 and y = f(x) for x between -1 and 0, and between the graphs of y = 1 and y = f(x) for x between 0 and 1.
Give an integral that corresponds to the area of this region.

however the answer to this is $\int_0^1 2(1-x^3)dx$

if I graph this i get a just an interval of 0 to 1

 

[CaptainBlack]

A region is bounded between the graphs of y = -1 and y = f(x) for x between -1 and 0, and between the graphs of y = 1 and y = f(x) for x between 0 and 1.
Give an integral that corresponds to the area of this region.

however the answer to this is $\int_0^1 2(1-x^3)dx$

if I graph this i get a just an interval of 0 to 1

Please post the entire question, preferably as asked.

In particular what is \(f(x)\)? Presumably \(f(x)=x^3\), but I am only guessing.

(If it is as I guess consider just one part of the graph and use symmetry to get the area for the other)

CB

 

[karush]

Let $f(x)= x^3$

A region is bounded between the graphs of $y = -1$ and $y = f(x)$ for $x$ between $-1$ and $0$, and between the graphs of $y = 1$ and $y = f(x)$ for x between $0$ and $1$. Give an integral that corresponds to the area of this region.

OK, sorry, I left out the $f(x)$, this is exactly the way it was given, however not sure why they split the interval

 

[soroban]

Hello, karush!

I don't suppose you made a sketch . . .

A region is bounded between the graphs of: $y = \text{-}1,\;y = x^3$ on $[\text{-}1,0]$
. . and between the graphs of: $y = 1,\;y = x^3$ on $[0,1]$.
Give an integral that corresponds to the area of this region.

                |
                |          *
                |
               1+ - - - - *
                |::::::::*|
                |::::::*  |
     -1         |:::*     |
  ----+---------+---------+----
      |     *:::|         1
      |  *::::::|
      |*::::::::|
      * - - - - +
              -1|
     *          |
                |

There are two sections to consider.
Since the two sections are symmetric (have equal areas)
. . we have find the area of one section and double.

$$A \;=\;2\int^1_0(1-x^3)\,dx$$Why did they split the region?

Evaluate: .$$\int^1_{\text{-}1}(1-x^3)\,dx$$. . and tell me what you get.

 

[karush]

ok so if we take the full interval of -1 to 1 they cancel to zero but since the areas are = at the given intervals then the area of the -1 to 1 is just double or 2 times.

https://www.physicsforums.com/attachments/511

thanks and I will more careful with the OP

 

[CaptainBlack]

ok so if we take the full interval of -1 to 1 they cancel to zero but since the areas are = at the given intervals then the area of the -1 to 1 is just double or 2 times.

https://www.physicsforums.com/attachments/511

thanks and I will more careful with the OP

The integral of \(f(x)=x^3\) from \(-1\) to \(1\) is indeed zero, but that is not what you are asked to find. You are asked to find the area of a region, and if you look at soroban's post you will see a sketch of that region.

Also the way the question is worded you are not dealing with signed areas, so if it had specified the area in your diagram it would have been asking for:

\[A=\int_{-1}^0 (0-x^3)\; dx + \int_{0}^1 (x^3-0)\; dx=2 \int_{0}^1 (x^3-0)\; dx\]

CB