Definite Integration

**gp3** · February 9th 2010, 09:36 PM

hi guys, i dont know where to place this integration question under, so i guess i'lljust place it here. I cant get the answer after many tries..

Find the area bounded by the given curves:

y=4-x square ; y=0, x=0, x=4

the answer given to me was 16. but I could only get -15.99 yes the ans I got is negative.

i solved by coming out with the table, and then integrate the equation by sub in the limit 4 and 0.

x 0 1 2 3 4
y 4 3 0 -5 -12

anyone help pls? thanks a lot!

I have end of semester exam coming!

**Soroban** · February 10th 2010, 06:28 AM

Hello, gp3!

Find the area bounded by: . $y\,=\,4-x^2,\;y=0,\;x=0,\;x=4$

The answer given to me was 16.

I believe the area looks like this:

Code:

                  |
                 4*..
              *   |:::*.      4
      -----*------+------*----+----
         *        |      2 *:::
        *         |         *:: 
                  |         ':: 
       *          |          *: 
                  |          ': 
                  |           : 
      *           |           * 
                  |

We want the area above the x-axis from 0 to 2:

. . $\int^2_0(4-x^2)\,dx \;\;=\;\;4x - \tfrac{1}{3}x^3\,\bigg]^2_0 \;\;=\;\;\left(8 - \tfrac{8}{3}\right) - (0-0) \;\;=\;\;\frac{16}{3}$

We want the area below the x-axis from 2 to 4:

. . $\int^4_2(x^2-4)\,dx \;\;=\;\;\tfrac{1}{3}x^3-4x\,\bigg]^4_2 \;\;=\;\;\left(\tfrac{64}{3} - 32\right) - \left(\tfrac{8}{3} - 8\right) \;\;=\;\;\tfrac{16}{3} - \left(-\tfrac{16}{3}\right) \;\;=\;\;\frac{32}{3}$

Total area: . $\frac{16}{3} + \frac{32}{3} \;\;=\;\;\frac{48}{3} \;\;=\;\;16$

**gp3** · February 10th 2010, 05:31 PM

hi thanks a lot!

but one thing i don't understand is that why the question have to give me y=0? I thought we just need to know the range of x axis which is 0 to 4.

Thread: Definite Integration

LinkBack

Thread Tools

Display