ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: keyan on August 15, 2012, 07:52:05 pm

Title: Wee integration question
Post by: keyan on August 15, 2012, 07:52:05 pm: Integral (with top value a and bottom value 0) f(x)dx = a

Then 2*integral (5a,0) (f(x/5) +3)dx =?

How do i go about answering this one? Also (5a,0) denotes the upper and lower values on the integral sign.
Title: Re: Wee integration question
Post by: DarkHorse on August 15, 2012, 08:05:01 pm: Tip: It's a dilation mate. :)
Title: Re: Wee integration question
Post by: keyan on August 15, 2012, 08:06:53 pm: Hah sorry forgot to write the proper question! It's f(x/5)
Title: Re: Wee integration question
Post by: BubbleWrapMan on August 20, 2012, 06:29:00 pm: Firstly

$2\int_{0}^{5a}(f(\frac{x}{5})+3)dx = 2(\int_{0}^{5a}f(\frac{x}{5})dx+\int_{0}^{5a}3dx)$

Note that f(x/5) is just f(x) dilated by a factor of 5 parallel to the x-axis, and the terminal 5a is also essentially a result of the dilation of 5 (try drawing a picture). So that part is just 5 times the area under the first graph, i.e.:

$\int_{0}^{5a}f(\frac{x}{5})dx = 5\int_{0}^{a}f(x)dx=5a$

As for the next part:

$\int_{0}^{5a}3dx =3(5a)-3(0) = 15a$

So putting it all together you should get:

$2(\int_{0}^{5a}f(\frac{x}{5})dx+\int_{0}^{5a}3dx)=2(5a+15a) = 40a$

Hope this helps (and that I'm right...).
Title: Re: Wee integration question
Post by: FlorianK on August 20, 2012, 07:12:19 pm: Climb you forgot tha it's twice the integral.
Additionally i think that this step isn't completly correct (maybe I'm wrong)

$\int_{0}^{5a}f(\frac{x}{5})dx = 5\int_{0}^{a}f(x)dx=5a$

Where do you get you factor 5 from?

Quote from: keyan on August 15, 2012, 07:52:05 pm
Integral (with top value a and bottom value 0) f(x)dx = a

Then 2*integral (5a,0) (f(x/5) +3)dx =?

How do i go about answering this one? Also (5a,0) denotes the upper and lower values on the integral sign.
This would be my solution (correct me if I'm wrong)

2*integral (5a,0) (f(x/5) +3)dx
=2* integral (5a,0) (f(x/5)) dx + 2* integral (5a,0) (3) dx
=2* integral (5a,0) (f(x/5)) dx + 30a
=2* intergal (a,0) (f(x)) dx + 30a
=2a + 30a
=32a
Title: Re: Wee integration question
Post by: BubbleWrapMan on August 20, 2012, 07:16:57 pm: Urgh yeah forgot the 2, I should get 40a.

As for how I got the factor, draw a diagram, it's hard to explain in words.

EDIT: Actually I'll try doing it this way

$\int_{0}^{5a}f(\frac{x}{5})dx = 5F(\frac{5a}{5})-5F(\frac{0}{5})<br />= 5(F(a)-F(0))<br />= 5\int_{0}^{a}f(x)dx<br />= 5a$

where F(x) is an antiderivative of f(x)

40a is the answer according to VCAA (it was on the 2010 exam).
Title: Re: Wee integration question
Post by: FlorianK on August 20, 2012, 10:44:34 pm: yeah, i get it now.

2*integral (5a,0) (f(x/5) +3)dx
=2* integral (5a,0) (f(x/5)) dx + 2* integral (5a,0) (3) dx
=2* integral (5a,0) (f(x/5)) dx + 30a
=10* intergal (a,0) (f(x)) dx + 30a
=10a + 30a
=40a