Checking my answer!

[Lizzie]

The problem:
Estimate the area under the graph of f(x)=25-x² from x=0 to x=4 using four approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or and overestimate?

My answer:
A=180 units² , overestimate

Just making sure I did it right ! I graphed it as well, but can't show the graph. Thanks!

 

[galactus]

Hey Lizzie:

Using the right endpoint method here is the computations:

n=4, \(\displaystyle \Delta(x)\)=4-0=4

\(\displaystyle \frac{4}{4}=1\)

\(\displaystyle 25-1^{2}=24\)

\(\displaystyle 25-2^{2}=21\)

\(\displaystyle 25-3^{2}=16\)

\(\displaystyle 25-4^{2}=9\)

\(\displaystyle (1)(24+21+16+9)=70\)

Here's a graph of your Riemann sum:

As you can see, there is an underestimate of the actual area of \(\displaystyle \frac{236}{3}=78.666......\)

 

[Lizzie]

ok, would that all change if I had went above the graph? or am i supposed to go under it?

 

[Lizzie]

Oh no, lol, I see what I did, right endpoints!!! OMG, what an idiot! I really ahve to start paying more attention, lol. Thanks so much!

 

[galactus]

The problem you posted said to estimate the area UNDER the graph. There's a whole lot of area outside of the graph.

 

[Lizzie]

lol, very true

Alrighty then, I've got it and understand what I did wrong! Thanks!!

 

[Gene]

You have implied that you have a TI-83 or TI-84 For verification on this problem:
Push y=
Enter the equation 25-x²
Push window
enter X_min=-7
enter X_max=7
enter X_scl=1
enter Y_min=-20
enter Y_max=30
enter Y_scl=1
{These would change with a different equation. You have to experiment with different values till you get a "nice" picture.}
Push Graph
Push 2nd then Trace to get CALC
pick #7
Enter 0 for the lower limit.
Enter 4 for the upper limit.
It tells you the integral of the equation is 78.666667 so your answer should be close to that.