Computing Area Between Curves/Lines Part II

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Hckyplayer8

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I browsed my book but I feel like I didn't really see any examples that fit the problem above.

Is there anything special about computing the integral of three functions as opposed to two?

I plotted the functions and found their intersections at [-4,2],[0,0], and [2,8].

Is the next step just finding the definite integral bounded by x=-4, x=2 of all three functions?
 
Hckyplayer8 said:
View attachment 14996
I browsed my book but I feel like I didn't really see any examples that fit the problem above.

Is there anything special about computing the integral of three functions as opposed to two?

I plotted the functions and found their intersections at [-4,2],[0,0], and [2,8].

Is the next step just finding the definite integral bounded by x=-4, x=2 of all three functions?
Click to expand...
It is the same as before. You will have to find the area between two curves. The thing is that it may not always be the same two curves over the whole region. Draw the graphs, as you did, and just realize that the sum of the areas will equal the total area.
 
Split the -4, 2 interval into 2 segments, where the area is bounded by only 2 functions.
 
Consider the following diagram:

fmh_0094.png

For the area shaded in red \(A_R\):

[MATH]A_R=\int_{-4}^0 (x+6)-\left(-\frac{x}{2}\right)\,dx=?[/MATH]
For the area shaded in green \(A_G\):

[MATH]A_G=\int_0^2 (x+6)-(x^3)\,dx=?[/MATH]
And then the total area \(A\) is:

[MATH]A=A_R+A_G=?[/MATH]
 
Thank you all for the help. Dividing the desired area into two segments by x=0 makes sense. Should have seen it.

AR = 36 and AG = 10 for a total of 46 units squared.

Thanks again all!
 
Looking at \(A_R\) as a triangle of base 6 and height 4, we should find it is 12:

Let's check the integration:

[MATH]A_R=\int_{-4}^0 (x+6)-\left(-\frac{x}{2}\right)\,dx=\int_{-4}^0 \frac{3}{2}x+6\,dx=\left[\frac{3}{4}x^2+6x\right]_{-4}^0=0-(-12)=12\quad\checkmark[/MATH]
The green area is correct. :)
 
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