Area Under the Curve

[Jason76]

Find the area under the curve for \(\displaystyle y = 4 - x^{2}\) for \(\displaystyle [-2, 2]\)

\(\displaystyle \int_{-2}^{2} 4 - x^{2}\)

\(\displaystyle \rightarrow 4x - \dfrac{x^{3}}{3}\) evaluated at \(\displaystyle -2\) (lower bound) and \(\displaystyle 2\) (upper bound)

\(\displaystyle [4(2) - \dfrac{(2)^{3}}{3}] - [4(-2) - \dfrac{(-2)^{3}}{3}] = \frac{32}{3} = 10.7\) Is this correct or on the right path?

 

[HallsofIvy]

\(\displaystyle \int_{-2}^{2} 4 - x^{2}\)

This should be \(\displaystyle \int_{-2}^2 4- x^2\)dx

The "dx" is important.

\(\displaystyle \rightarrow 4x - \dfrac{x^{3}}{3}\) evaluated at \(\displaystyle -2\) (lower bound) and \(\displaystyle 2\) (upper bound)

\(\displaystyle [4(2) - \dfrac{(2)^{3}}{3}] - [4(-2) - \dfrac{(-2)^{3}}{3}] = \frac{32}{3} = 10.7\) Is this correct or on the right path?

Well, strictly speaking \(\displaystyle \dfrac{32}{3}\) is NOT equal to 10.7. \(\displaystyle \dfrac{32}{3}\) is the correct answer, 10.7 is not.

 

[Jason76]

This should be \(\displaystyle \int_{-2}^2 4- x^2\)dx

The "dx" is important.


Well, strictly speaking \(\displaystyle \dfrac{32}{3}\) is NOT equal to 10.7. \(\displaystyle \dfrac{32}{3}\) is the correct answer, 10.7 is not.

Why is that so:?

 

[JeffM]

Why is that so:?

I would prefer to say that \(\displaystyle \dfrac{32}{3}\) is the exact answer whereas \(\displaystyle 10.7\) is an approximate answer.

So, being careful, I would say \(\displaystyle \dfrac{32}{3} \approx 10.7,\ not\ \dfrac{32}{3} = 10.7.\)