Author Topic: Integration Problem (Read 1150 times) Tweet Share

BakerDad12 · « **on:** May 01, 2020, 06:56:25 pm »

Hey. I'm confused about a property of integrals. I've attached a picture below to demonstrate the type of question I am talking about. In that question, if you take the 1/2 out, you get a different answer to if you didn't because 1 and -1 cancel each other out. This isn't correct, so why can't you take the 1/2 out? What is the rule for this?

Einstein_Reborn_97 · « **Reply #1 on:** May 01, 2020, 07:10:13 pm »

Quote from: BakerDad12 on May 01, 2020, 06:56:25 pm

Hey. I'm confused about a property of integrals. I've attached a picture below to demonstrate the type of question I am talking about. In that question, if you take the 1/2 out, you get a different answer to if you didn't because 1 and -1 cancel each other out. This isn't correct, so why can't you take the 1/2 out? What is the rule for this?

It seems like you haven't attached any picture.

BakerDad12 · « **Reply #2 on:** May 01, 2020, 09:27:08 pm »

Whoops, sorry! It should be fixed now.

Einstein_Reborn_97 · « **Reply #3 on:** May 01, 2020, 10:48:39 pm »

If you look closely, the term -1 is also part of the integration. So, yes, you can take out 1/2 but you need to make sure you take it out from the 1 as well...
$∫(\frac{1}{2}(1+x)-1)dx$
$=\frac{1}{2}∫(1+x-2)dx$
$=\frac{1}{2}∫(x-1)dx$
$=\frac{1}{2}(\frac{x^2}{2}-x)+C$
$=\frac{x^2}{4}-\frac{x}{2}+C$

Here's an alternative method:
$∫(\frac{1}{2}(1+x)-1)dx$
$=∫\frac{1}{2}(1+x)dx-∫1dx$
$=\frac{1}{2}∫(1+x)dx-∫1dx$
$=\frac{1}{2}(x+\frac{x^2}{2})-x+C$
$=\frac{x}{2}+\frac{x^2}{4}-x+C$
$=\frac{x^2}{4}-\frac{x}{2}+C$
Hope that helps!

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