Is it just me, or evaluating integrals is not the easist thing to do? Please help!

**theunforgiven** · December 4th 2012, 02:25 AM

This is my first time evaluating integrals on my own and, unfortunately, I just can't seem to get it. As easy as they may seem, integrals are giving me such a hard time, even the simplest ones are proving to be a challenge.

I need help evaluating the integral below:

I ended up with 15ln(-1)-15ln(-3)=DNE, but I'm almost certain I got it wrong.

**BobP** · December 4th 2012, 02:37 AM

You should have a list of standard integrals, if you haven't got one then get one. Later, it helps if you can commit them to memory.
The one you need here is $\int \frac{1}{x}\, dx = \log_{e}x + C.$

**Prove It** · December 4th 2012, 02:38 AM

Actually it's $\displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{|x|} + C \end{align*}$

**theunforgiven** · December 4th 2012, 02:41 AM

Originally Posted by Prove It

Actually it's $\displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{|x|} + C \end{align*}$

That's what I was about to say. Where do I take it from there though? I know that one would be $15ln{|x|} + C$ but the numbers, -1 and -3, on that symbol are confusing the hell out of me!
If I knew what to do, I wouldn't be asking...

**Prove It** · December 4th 2012, 02:49 AM

The absolute value turns anything negative positive.

**theunforgiven** · December 4th 2012, 02:52 AM

Originally Posted by Prove It

The absolute value turns anything negative positive.

Oh hold on a sec! Would the answer be:
$15ln|-1|-15ln|-3|=-16$ (approximated)
?

**BobP** · December 4th 2012, 03:14 AM

Originally Posted by Prove It

Actually it's $\displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{|x|} + C \end{align*}$

Agreed, lazy of me.

**Prove It** · December 4th 2012, 03:24 AM

Originally Posted by theunforgiven

Oh hold on a sec! Would the answer be:
$15ln|-1|-15ln|-3|=-16$ (approximated)
?

No. Like I said, absolute values turn anything negative positive. So $\displaystyle \begin{align*} 15\ln{|-1|} - 15\ln{|-3|} = 15\ln{(1)} - 15\ln{(3)} = 0 - 15\ln{(3)} = -15\ln{(3)} \end{align*}$

**theunforgiven** · December 4th 2012, 03:29 AM

Originally Posted by Prove It

No. Like I said, absolute values turn anything negative positive. So $\displaystyle \begin{align*} 15\ln{|-1|} - 15\ln{|-3|} = 15\ln{(1)} - 15\ln{(3)} = 0 - 15\ln{(3)} = -15\ln{(3)} \end{align*}$

And that's exactly what I did. I just didn't show that extra step where you put the positive values between parentheses. I did, however, turn them to positive and put them between parentheses when I plugged in the equation into my calculator.

But yeah, thank you so much!

**HallsofIvy** · December 4th 2012, 04:51 AM

Originally Posted by theunforgiven

Oh hold on a sec! Would the answer be:
$15ln|-1|-15ln|-3|=-16$ (approximated)
?

Yes, although -16.5 would be a better approximation.

One of the first things you should have learned about "definite integrals" (actually, the definition) is that if F is an anti-derivative of f (that is, if dF/dx= f) then $\int_a^b f(x)dx= F(b)- F(a)$ .

Actually, yes, finding anti-derivatives is, typically harder than differentiating- although evaluating definite integrals should not be more difficult than evaluating any functions. There is a general distinction in mathematics between "direct problems", where we are given a specific definition or formula, and "inverse problems" where we are to "reverse" some direct problem. For example, if I told you that $y= x^5- 3x^4+ 4x^2- 7x+ 9$ and ask you to find y when x= 3, that would be easy- just do the arithmetic in that formula. But if I told you that y= 3 and asked you to find x,, that would be a much more difficult problem- and there might be many solutions or none.

We learn, in introductory Calculus, a formula for the derivative of a function so finding a derivative can always go back to using that formula (though we can also use the many "theorems" for special cases derived from that formula). That's a direct problem. But an anti-derivative of f, say, is only defined as "a function whose derivative is f". We don't have a direct formula to use, we have to try to "remember" a function whose derivative is the given function. That's the "inverse" problem.

**theunforgiven** · December 4th 2012, 09:36 AM

Originally Posted by HallsofIvy

Yes, although -16.5 would be a better approximation.

One of the first things you should have learned about "definite integrals" (actually, the definition) is that if F is an anti-derivative of f (that is, if dF/dx= f) then $\int_a^b f(x)dx= F(b)- F(a)$ .

Actually, yes, finding anti-derivatives is, typically harder than differentiating- although evaluating definite integrals should not be more difficult than evaluating any functions. There is a general distinction in mathematics between "direct problems", where we are given a specific definition or formula, and "inverse problems" where we are to "reverse" some direct problem. For example, if I told you that $y= x^5- 3x^4+ 4x^2- 7x+ 9$ and ask you to find y when x= 3, that would be easy- just do the arithmetic in that formula. But if I told you that y= 3 and asked you to find x,, that would be a much more difficult problem- and there might be many solutions or none.

We learn, in introductory Calculus, a formula for the derivative of a function so finding a derivative can always go back to using that formula (though we can also use the many "theorems" for special cases derived from that formula). That's a direct problem. But an anti-derivative of f, say, is only defined as "a function whose derivative is f". We don't have a direct formula to use, we have to try to "remember" a function whose derivative is the given function. That's the "inverse" problem.

That helped immensely! Thank you

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