Wee integration question

[keyan]

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.

[DarkHorse]

Tip: It's a dilation mate. 

[keyan]

Hah sorry forgot to write the proper question! It's f(x/5)

[BubbleWrapMan]

Firstly



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.:



As for the next part:



So putting it all together you should get:



Hope this helps (and that I'm right...).

[FlorianK]

Climb you forgot tha it's twice the integral.
Additionally i think that this step isn't completly correct (maybe I'm wrong)



Where do you get you factor 5 from?

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

[BubbleWrapMan]

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



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

40a is the answer according to VCAA (it was on the 2010 exam).

[FlorianK]

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