Checkpoint Question

[Sanguinne]

Can anyone explain this question, I dont understand checkpoint's explanation. Q 89 from vcaa 2010

Let f be a differentiable function defined for all real x, where f(x)≥0 where x ∈ [0,a].

If = a, then 2 is equal to

EDIT: at the top of the second integral, the top part is suppose to be 5a without any gaps

[Zealous]

I had to think about this for a while before realising how to do it.

The trick (I think...) is that



and



are exactly the same.

What happens when you have ? Well, the graph is dilated by a factor of 5 along the x-axis.
And it happens to be that the integral, has an upper limit which is 5 times of the original f(x)

edit my image is wrong..

Spoiler

Have a look at this picture here:

Hopefully you can see, that the "integral" or "area" will be exactly the same.

So:



=           split the integral...

=           sub in 5a...

=           evaluate the integral of 3...

=

[edit]: making the math look pretty =)

[plato]

Sushi is correct in the explanation but the graph suggests a dilation away from both axes. In fact the dilation from f(x) to f(x/5) is only away from the y axis (parallel to the x axis).

Therefore the graph of f(x/5) should have the same height as the graph for f(x). That would then show an area five times greater than the original as Sushi has explained.

[Zealous]

Sushi is correct in the explanation but the graph suggests a dilation away from both axes. In fact the dilation from f(x) to f(x/5) is only away from the y axis (parallel to the x axis).

Therefore the graph of f(x/5) should have the same height as the graph for f(x). That would then show an area five times greater than the original as Sushi has explained.

Oh yeah oops, sorry I stuffed up the picture. I'll fix it soon.

[Sanguinne]

Thanks for the great explanation