# Thread: Finding the total area between the function and the x-axis

1. ## Finding the total area between the function and the x-axis

I'm supposed to find the area in the region between the function and the x-axis....

y = x^(1/3), [-1,8]

I divided it up into two regions (-1 to 0 -> A1 and 0 to 8 -> A2). I found A2 to be 12 and A1 to be |-3/4|. My teacher worked the problem on the review sheet and she got |-1| for A1. How is this possible?!

We both have the same equation:

A1 = (3/4)x^(4/3) evaluated from 0 to -1. Obviously x=0 will yield 0, and when I plug in -1 I get -3/4! Am I just completely missing something?

For the total area, I get 12.75 units, and my teacher got 13.

2. Originally Posted by tom ato
I'm supposed to find the area in the region between the function and the x-axis....

y = x^(1/3), [-1,8]

I divided it up into two regions (-1 to 0 -> A1 and 0 to 8 -> A2). I found A2 to be 12 and A1 to be |-3/4|. My teacher worked the problem on the review sheet and she got |-1| for A1. How is this possible?!

We both have the same equation:

A1 = (3/4)x^(4/3) evaluated from 0 to -1. Obviously x=0 will yield 0, and when I plug in -1 I get -3/4! Am I just completely missing something?

For the total area, I get 12.75 units, and my teacher got 13.
I'm with you. You'll notice that your function is symmetry about the origin so $A1 = \int_0^1 x^{1/3}\,dx = \left. \frac{3}{4} x^{4/3} \right |_0^1 = 3/4$ as you say. Further

$A2 = \int_0^8 x^{1/3}\,dx = \left. \frac{3}{4} x^{4/3} \right |_0^8= 12$

3. $A_1 = \int_{-1}^0 (0-x^{1/3})dx=\left[-\frac{3}{4}x^{4/3}\right]_{-1}^0=\frac{3}{4}$

4. Originally Posted by tom ato
My teacher worked the problem on the review sheet and she got |-1| for A1.
she forgot the factor $1/(n+1)$ perhaps.