OK, I actually need help on this one.

[Lizzie]

Now that I understand integrating (it was really easy once I figured it out), I have a problem that I need help with.

The problem:
The actual area (using the Fundamental Theorem of Calculus) for the function f(x)=16-x² is:

What I know so far:
ok, I know that the FTC is as follows -
\(\displaystyle \int^b_a{}f(x)dx= F(b)-F(a)\), if F is the antiderivative of f.
And
\(\displaystyle \frac{d}{dx}(\int^x_a{}f(t)dt)= f(x)\) if \(\displaystyle a\) is a constant.

So, I know the first step is to integrate 16-x², which I did on a previous post. This is \(\displaystyle 16x-\frac{x^3}{3}+c\).

I need help understanding this next bit, so any help would be appreciated. Once again, I would like to emphasize that I am not looking for the answer, just some help getting there. Thank you!

 

[stapel]

Evaluate at the upper limit "b".

Evaluate at the lower limit "a".

Subtract the latter from the former.

Eliz.

 

[Lizzie]

Oh duh! That is so obvious!!! *smacks self in head* What an idiot! lol. Thanks again Stapel! I can't believe I just did that...

 

[stapel]

It's "obvious" once you've done it once or twice. There will be any number of things in calculus that will be "obvious" after you see somebody do it once. That's just how it works. :wink:

Eliz.

 

[Lizzie]

I got -62...is that close? lol

 

[stapel]

Looking at the graph, does that seem reasonable? Considering your "by rectangles and endpoints" approximation, does that sound close?

Are you sure you subtracted in the right order? "F(2) - F(-2)", instead of "F(-2) - F(2)"? :wink:

Eliz.

 

[Lizzie]

hah, I did it backwards, that's what was wrong, my new answer was 12. . .

 

[Lizzie]

Ok, 12 doesn't seem right either...or am I just imagining it...?

 

[Lizzie]

Is there more to it than just this? Because this was a multiple choice and all of the possible answers are much higher....

 

[pka]

Do you understand this?
\(\displaystyle \L
\int_1^3 {\left( {16 - t^2 } \right)} dt = \left. {16t - \frac{{t^3 }}{3}} \right|_{t = 1}^{t = 3}\ \ = \left( {16(3) - \frac{{(3)^3 }}{3}} \right) - \left( {16(1) - \frac{{(1)^3 }}{3}} \right)\)
Well then \(\displaystyle \L
\int_a^x {\left( {16 - t^2 } \right)} dt = \left. {16t - \frac{{t^3 }}{3}} \right|_{t = a}^{t = x}\ \ = \left( {16(x) - \frac{{(x)^3 }}{3}} \right) - \left( {16(a) - \frac{{(a)^3 }}{3}} \right)\) .
Find the derivative with respect to x, \(\displaystyle \L
\left( {16(x) - \frac{{(x)^3 }}{3}} \right) - \left( {16(a) - \frac{{(a)^3 }}{3}} \right)\) and get \(\displaystyle \L
{\left( {16 - x^2 } \right)}\)

 

[stapel]

Note: You didn't include the actual limits on the integral. I assumed, from a previous post, that the limits were a = -2 to b = 2. The reply above assumes a = 1 to b = 3. The different limits will, of necessity, return different answers.

Eliz.

 

[Lizzie]

it was -2 and 2, sorry for not reposting that. Thanks both of you, I will continue working with that.

 

[Lizzie]

Ok, so I take what pka said, which is what I had, and sub in 2 and -2, which I did.

\(\displaystyle \L\int^2_{-2}{}16x-\frac{x^3}{3}\)
. . .
\(\displaystyle \L\int{}(16(2)-\frac{(2)^3}{3}) - \int{}(16(-2)-\frac{(-2)^3}{3})\)
. . .
\(\displaystyle \L\int{}(32-\frac{8}{3}) - \int{}(-32-\frac{-8}{3})\)
. . .
\(\displaystyle \L{}32-\frac{8}{3}+32+\frac{8}{3}\)
. . .
I keep getting different answers everytime I do it....BLAH!

 

[Lizzie]

...Sorry guys

 

[Lizzie]

ok, here are the actual choices.

A) \(\displaystyle \frac{176}{3}\)
B) \(\displaystyle \frac{88}{3}\)
C) \(\displaystyle 60\)
D) cannot be determined

I highly doubt that it's D, but I can't seem to get this... I get the first bit of it now, it's just the last bit, now.

 

[Unco]

G'day, Lizzie.

Your work is great.

Ok, so I take what pka said, which is what I had, and sub in 2 and -2, which I did.

\(\displaystyle \L \int^2_{-2} 16 \, - \, x^2 \, dx\)

\(\displaystyle = \, \L\left[16x-\frac{x^3}{3}\right]^2_{-2}\)

\(\displaystyle = \, \L\left(16(2)-\frac{(2)^3}{3}\right) \, - \, \left(16(-2)-\frac{(-2)^3}{3}\right)\)

\(\displaystyle = \, \L\left(32-\frac{8}{3}\right) \, - \, \left(-32-\frac{-8}{3}\right)\)

\(\displaystyle = \, \L\left(32-\frac{8}{3}\right) \, - \, \left(-32+\frac{8}{3}\right)\)

\(\displaystyle = \, \L32-\frac{8}{3}+32+\frac{8}{3}\) <- See? That 8/3 should be negative.

 

[Lizzie]

Ok, so if it's negative, then
\(\displaystyle 32-\frac{8}{3}+32-\frac{8}{3}=62-\frac{16}{3}=\frac{170}{3}\)
That's still not one of my answers... am I still doing seomthing wrong?

 

[Unco]

32 + 32 = 62 ?

 

[Lizzie]

OMG, I keep doing that, lol. I accidentally did that about 3 times and caught myself every time but this one! Thanks