Final Exam HELP

[jack_chay]

I have some questions that my teacher has given me in practise exams.

integral of 5 to 1 ( f(x) ) dx = 6

then

2 times integral of 5 to 1 ( f(x) + 3) dx equals:

thanks in advance

[Alwin]

I have some questions that my teacher has given me in practise exams.

integral of 5 to 1 ( f(x) ) dx = 6

then

2 times integral of 5 to 1 ( f(x) + 3) dx equals:

thanks in advance

Consider f(x) = 3/2

Try it and see what you get!

[rhinwarr]

Use the rule that:
Integral of f(x) + c is equal to the integral of f(x) + integral of c

Therefore:
2 times integral of 5 to 1 ( f(x) + 3) dx equals:
2 times integral of 5 to 1 f(x) dx + integral of 5 to 1 (3) dx
=2 times 6 times (15 - 3)
=12 times 12
=144

[jack_chay]

um... but the only options are 9, 15, 18, 24 and 36, there are no 144??

[jack_chay]

don't worry I got it, it's 36, thank you

[rhinwarr]

Woops, I times it instead of added it.
2 x (6 + 12) = 36

[Homer]

was this from MAV 2012? if yes then it is 36

[jack_chay]

all the jibberish...

i have no idea what the question is asking...

thank you in advance

[SocialRhubarb]

Haha, where's this from?

I'm pretty sure it's just asking for the integral of from 0 to 4.

[b^3]

What the question is saying is that we're taking the function and over the interval , we're going to divide the area under this curve up into rectangles, in this case rectangles. The width of a rectangle here will be , so you can interpret it as a 'small change' in , or that the width of the rectangles is a small portion of our interval. Now we're going to look at what happens when we sum the area of these rectangles, but make the rectangle width go to zero. That is as . What we end up with is just the area under the curve in that interval, which as SocialRhubarb has said, will just be the definite integral of the function over that interval.

Hope that makes sense.

[Phy124]

Ok so remember when you first learnt about integration you did it by making rectangles under the graph and finding the area of these rectangles?

Well when you did that, maybe you had 3 rectangles or 4 or 5 or 6 etc.

This question is stating that we are making rectangles and summing the area of these rectangles (with width i.e. the right value of of the rectangle ,, minus the left value of of the rectangle, , and height i.e. the value of to the right squared).

The value is practically infinite as we are finding the sum at which the width of the rectangles is approaching zero.

This is the underlying principle which integration is based on but not many people seem to know much about. You'd be much more familiar with the form , no?

[jack_chay]

how do you know it is o to 4, i thought it was 1 to 4...

[Phy124]

The key parts to that are;

...

... ...

So it's saying you're doing it over the interval [0,4] and if you don't pick it up from that the second part I quoted is saying that , the first point you're taking (that is, the left value of the left most rectangle) is

[jack_chay]

oh yeah.. hehehe sorry thank you