How do you find the area of the region bounded by the graph of #f(x)=1-x^2# and the #x#-axis on the interval #[-1,1]# ?

Question

Scarlett Allison · Accepted Answer

This is a definite integral problem. #int_-1^1(1-x^2)dx# #=[x-x^3/3]_-1^1# #=[(1)-(1)^3/3-((-1)-(-1)^3/3)]# #=[1-1/3-(-1-(-1)/3)]# #=[1-1/3-(-1+1/3)]# #=[1-1/3+1-1/3]# #=[2-2/3]# #=[6/3-2/3]# #=[4/3]=1.3333-># Solution

Aurora Alvarado · Answer

undefined To find the area of the region bounded by the graph of $ f(x) = 1 - x^2 $ and the x-axis on the interval $[-1, 1]$, follow these steps:

1. Determine the x-values where $ f(x) = 0 $ to find the bounds of integration.

$ 1 - x^2 = 0 $
   
   $ x^2 = 1 $
   
   $ x = \pm 1 $

2. Set up the definite integral to find the area:

$$ \text{Area} = \int_{-1}^{1} |f(x)| \, dx $$

3. Substitute the function $ f(x) = 1 - x^2 $ into the integral:

$$ \text{Area} = \int_{-1}^{1} |1 - x^2| \, dx $$

4. Evaluate the integral.

Since the function is symmetric about the y-axis, you can split the integral into two parts:

$$ \text{Area} = 2 \int_{0}^{1} (1 - x^2) \, dx $$

5. Integrate $ (1 - x^2) $ from $ 0 $ to $ 1 $:

$$ \text{Area} = 2 \left[ x - \frac{x^3}{3} \right]_{0}^{1} $$

$$ \text{Area} = 2 \left[ \left(1 - \frac{1}{3}\right) - \left(0 - 0\right) \right] $$

$$ \text{Area} = 2 \left( \frac{2}{3} \right) $$

$$ \text{Area} = \frac{4}{3} $$

Therefore, the area of the region bounded by the graph of $ f(x) = 1 - x^2 $ and the x-axis on the interval $[-1, 1]$ is $ \frac{4}{3} $ square units.