# Quick question on why I got something wrong

• Nov 15th 2011, 11:30 PM
NeedDirection
Quick question on why I got something wrong
I was given $\displaystyle \int^4_0 f(x)\,dx = 4$

I was asked to evaluate the following:
$\displaystyle \int^4_0 (3*f(x)+1)\,dx$

What I simply did was took $\displaystyle 3*4+1=13$ and I was marked down a point. How is this incorrect? I'm sure it's something stupid I am not seeing but I can't seem figure it out!
Thanks
• Nov 16th 2011, 12:02 AM
Opalg
Re: Quick question on why I got something wrong
Quote:

Originally Posted by NeedDirection
I was given $\displaystyle \int^4_0 f(x)\,dx = 4$

I was asked to evaluate the following:
$\displaystyle \int^4_0 (3*f(x)+1)\,dx$

What I simply did was took $\displaystyle 3*4+1=13$ and I was marked down a point. How is this incorrect? I'm sure it's something stupid I am not seeing but I can't seem figure it out!
Thanks

What is $\displaystyle \int_0^41\,dx$ ? (Wink)
• Nov 16th 2011, 12:03 AM
FernandoRevilla
Re: Quick question on why I got something wrong
Quote:

Originally Posted by NeedDirection
What I simply did was took $\displaystyle 3*4+1=13$ and I was marked down a point. How is this incorrect? I'm sure it's something stupid I am not seeing but I can't seem figure it out!

The solution is not $\displaystyle 13$ :

$\displaystyle \int_0^4(3f(x)+1)dx=\int_0^43f(x)dx+\int_0^41dx=$

$\displaystyle 3\int_0^4 f(x)+[x]_0^4=3\cdot 4+(4-0)=16$

Edited: Sorry, I didn't see Opalg's post.
• Nov 16th 2011, 12:27 AM
NeedDirection
Re: Quick question on why I got something wrong
Just as I expected, something simple I couldn't see!
Thanks guy's!