Question

Calculus help please?

Answer 1

sorry i dont know any calculus

Answer 2

@jhonyy9

Answer 3

Do you know of any which you think are definitely true?

Answer 4

No, I have no recollection on how to do this problem... @bobo-i-bo

Answer 5

Do you know how to integrate?

Answer 6

Yeah

Answer 7

Okay, then let f(x)=1 and g(x)=1 as well. Then try to apply the integrals and you should see which ones are incorrect which will reduce the answers. Get what I mean? :)

Answer 8

Hmmm... not sure I'm catching on

Answer 9

1 is right I think

Answer 10

So for example, if f(x)=1 and g(x)=1, for the first one:
\[\int^a_b f(x)g(x)dx=\int^a_b (1* 1) dx=b-a\]
but
\[ ( \int^a_bf(x)dx)(\int^a_b g(x)dx)=(\int^a_b 1dx)(\int^a_b 1 dx)=(b-a)(b-a)\]
So it is not always true that:
\[\int^a_b f(x)g(x)dx=(\int^a_b f(x)dx)(\int^a_b g(x)dx)\]
since it's not true when f(x)=1 and g(x)=1.

Answer 11

Do you see? :)

Answer 12

Ok yes

Answer 13

Are u the real loren beech

Answer 14

Integration is a linear operation,
so it follows the two rules for linearity,\[\large\rm L(x+y)=L(x)+L(y)\]\[\large\rm L(ax)=aL(x),\qquad\forall a\in\mathbb R\]

So we can see that number is certainly true! :)\[\large\rm \int\limits f(x)+g(x)~dx\quad=\quad \int\limits f(x)dx+\int\limits g(x)dx\]The third one I'm not so sure about :d still thinking...

Answer 15

number 2 is certainly true*

Answer 16

Thank you! Tricky number 3 ugh

Answer 17

For number 3, let's just take a look at a simple f(x) that satisfies the requirements, and see if we can break it.

So like, f(x)=x^2.

\(\large\rm \int_a^b \sqrt{x^2}dx=\int_a^b x dx=\frac{1}{2}x^2|_a^b=\frac{1}{2}(b^2-a^2)\)

If we look at it the other way,
\(\large\rm \sqrt{\int_a^b x^2dx}=\sqrt{\frac{1}{3}x^3|_a^b}=\sqrt{\frac{1}{3}(b^3-a^3)}\)

So ummm... I think we would say that three is false...
The square root function doesn't behave nicely when removed from the integration process.

Hmm.

Answer 18

@zepdrix correct ^_^