Integration (1 Viewer)

[independantz]

Find the area bounded by the graph of the given function and the coordinate axes:
y=(x+2)^3

The answer is 4 units^2

What exactly are the boundaries for this question ??

This question is from the year 11 cambridge book, 11E, Q10


Thanks in advance

 

[watatank]

i'm pretty sure the limits are (-2,0) but you should draw a graph just to make sure

 

[Slidey]

That's what you have to find.

The y coordinate axis is of form x=0. So sub x=0 into the function:
y=(0+2)^3=8
So one boundary is (0,8) - but you only really needed the information x=0, since the terminals of integration (the numbers at the top and bottom of the f-like integral sign) are x values.
The x axis is of form y=0, so find the roots:
0=(x+2)^3
x=-2
Other boundary is: (-2,0).

Draw quick sketch and you'll see the area you need to find is entirely above the x axis and hence positve, so you don't need to worry about absolute values and stuff.

Integrate y from -2 to 0 w.r.t. x:

Int (x+2)^3 dx from -2 to 0
either do the substitution in your head or write it out: u=x+2, du=dx
So the answer is (1/4)*[(x+2)^4] from -2 to 0 or: (1/4)*(16-0)=4 square units.