Find the area between the curve and the x-axis

[sigma]

I need help with 2 of these types of questions. I'm also asked to sketch the graph.

\(\displaystyle \
\L\
\begin{array}{l}
1.{\rm Find the area of: }y = x^2 - 4x \\
{\rm and so: }y = x^2 - 4x \\
0 = x(x - 4) \\
x = 0,4 \\
\int_0^4 {x^2 - 4x{\rm dx} = \frac{{x^3 }}{3}} - \left. {2x^2 } \right|_0^4 = \left( {\frac{{4^3 }}{3} - 2(4)^2 } \right) - \left( {0 - 0} \right) \\
= \frac{{64}}{3} - \frac{{96}}{3} = - \frac{{32}}{3} \\
\end{array}
\\)
Is this right? For this question, I'm not sure what the graph is suppose to look like. I know its a parabula that opens up but I'm not 100% sure what to do with the -4x part.


\(\displaystyle \
\L\
\begin{array}{l}
2.{\rm Find the area of: }y = 9 - x^2 {\rm between the lines }x = - 1{\rm and }x = 5 \\
{\rm and so: }\int_{ - 1}^5 {9 - x^2 } {\rm dx} = 9x - \left. {\frac{{x^3 }}{3}} \right|_{ - 1}^5 = \left( {45 - \frac{{5^3 }}{3}} \right) - \left( { - 9 + \frac{1}{3}} \right) = \frac{{10}}{3} + \frac{{26}}{3} = \frac{{36}}{3} = 12 \\
\end{array}
\\)
I have the graph (if its right):


For this question, I want to know if you just go straight to the integral from -1 to 5 (which I've done) and don't take into account factoring the original equation of y, which would give you 3 and -3. Like the first quesiton, I would factor it, find the critical points, and intergrate with that if they don't specificaly ask for any other area between the lines like this question?

 

[stapel]

1) "Find the area of" usually means "find the positive integral of", or else "Find the area below the curve and above the x-axis". In other words, usually the "area" is already positive (being above the x-axis), or you're supposed to make it positive.

For this exercise, are you sure you're supposed to be finding negative areas? That is, are you sure you're supposed to be finding just the value of the integral, rather than its absolute value (and thus the positive measurement of "area" under the x-axis)?

Texts can go either way, so you'll need to verify, from worked examples and/or solutions to other exercises, which your book means.

2) Same question as before, especially since this one has both positive and negative areas.

Eliz.

 

[sigma]

I don't think the question is specifying the area just above the x-axis, but between the curve and the x-axis, which can be above or below. It can be either negative or positive as long as its between the curve and the x-axis. The curve can go below the x-axis, meaning the area will be negative but I'm not 100% certain. We just started these a few days ago.

Why would question 2 have both positive and negative areas? According to the graph (if its right) the area will be positive because its above the x-axis. But again, I don't know if that is right. All the question states is: Find the area between the curve and the x-axis.

 

[sigma]

\(\displaystyle \
\L\
\begin{array}{l}
1.{\rm Find the area of: }y = x^2 - 4x \\
{\rm and so: }y = x^2 - 4x \\
0 = x(x - 4) \\
x = 0,4 \\
\int_0^4 {x^2 - 4x{\rm dx} = \frac{{x^3 }}{3}} - \left. {2x^2 } \right|_0^4 = \left( {\frac{{4^3 }}{3} - 2(4)^2 } \right) - \left( {0 - 0} \right) \\
= \frac{{64}}{3} - \frac{{96}}{3} = - \frac{{32}}{3} \\
\end{array}
\\)

Yea something does not seem right here. The answer should be positive, not negative. I know with areas below the x-axis (above the curve), you have to take the absolute value or negate it somehow. What part exactely should I be negating though? I still don't know what the graph is suppose to look like. Can somebody help me out here?

 

[galactus]

 

[sigma]

I almost had that graph. What would the curve look like before it passes through zero? So the area can be negative then?

 

[stapel]

So the area can be negative then?

That's what I was asking: How does your text define "area"?

Thank you.

Eliz.

 

[sigma]

Here's what my text says:
\(\displaystyle \
\L\
\begin{array}{l}
{\rm Definition of area by integration} \\
area = \int_a^b {\left| {f(x)} \right|} dx \\
\end{array}
\\)

Then they show an example of using area below the x-axis
\(\displaystyle \
\L\
\begin{array}{l}
y = x^2 - 4 \\
= - \int_{ - 2}^2 {(x^2 - 4)dx{\rm }\left( {{\rm or}\left| {\int_{ - 2}^2 {} (x^2 - 4)dx} \right|{\rm in this case}} \right)} \\
= - \left[ {\left. {\left( {\frac{{x^3 }}{3} - 4x} \right)} \right|_{ - 2}^2 } \right] = - \left[ {\left( {\frac{8}{3} - 8} \right) - \left( { - \frac{8}{3} + 8} \right)} \right] = \frac{{32}}{3} \\
\end{array}
\\)

I guess this makes sense now.

 

[sigma]

It does make sense to me now. I do have to negate the integral if the curve is below the x-axis. So my answer for 1 is the same but just positive instead.

Other than that, did I do the questions correctly?