# Integral

Discussion in 'Math' started by Lightfire, Oct 1, 2013.

1. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21

Hello fellows,

Just like what I did in my derivative question, I'd also like to ask if my answer is correct in integrating.

Given of

$\int_2^3 2x^{3}+11x^{2}+31x.dx$

$=\int 2x^{3}+\int 11x^{2}+\int 31x$

$=\frac {2x^{3+1}}{3+1}+\frac{11x^{2+1}}{2+1}+\frac{31x^{1+1}}{1+1}$

$=\frac {2x^{4}}{4}+\frac{11x^{3}}{3}+\frac{31x^{2}}{2}$

$=\frac {x^{4}}{2}+\frac{11x^{3}}{3}+\frac{31x^{2}}{2}$

$=\frac {3x^{4}+22x^{3}+93x^{2}}{6}$$\]_2^3$

At x = 2

$=\frac {3(2^{4})+22(2^{3})+93(2^{2})}{6}$

$=\frac {3(16)+22(8)+93(4)}{6}$

$=\frac {48+176+372}{6}$

$=\frac {576}{6}$

At x = 3

$=\frac {3(3^{4})+22(3^{3})+93(3^{2}}{6}$

$=\frac {3(81)+22(27)+93(9)}{6}$

$=\frac {243+594+837}{6}$

$=\frac {1674}{6}$

Subtracting:

$\frac {1674}{6}-\frac {596}{6}$

$=\frac {1078}{6}$

$=179.\overline{6}$

Thank you very much!

Last edited: Oct 1, 2013
2. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
That looks good to me. Just be careful with your rounding at the end. It is 179.66.....7, so if rounding to one decimal place it would be 179.7.

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3. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Is 179.6... different from 179.666...?

4. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
When I did the problem out the answer I got 179.6666....etc.

1078/6 is equal to 539/3, which is equal to 179.666(repeated). Therefore it should be rounded to 179.7.

Remember, it's generally a better practice not to round until the very end. Otherwise you can lose a lot of precision.

5. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Yup, it should be rounded as 179.7 (to the nearest tenth). But isn't 179.6... ("..." is ellipsis) means that the 6 after the decimal point repeats forever?

6. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
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I just noticed the ellipsis and realized that's probably what you meant. Anyway, your answer is correct

7. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Note
It is important to distinguish between the indefinite integral which has no limits

${\int {2x} ^3} = \frac{{2{x^4}}}{4} + C$

and the definite integral which has two limits

$\int\limits_a^b {2{x^3}} = \left[ {\frac{{2{x^4}}}{4}} \right]_a^b = \frac{{2{b^4}}}{4} - \frac{{2{a^4}}}{4}$

Note the constant in the indefinite integral and try to avoid writing lines like your second, third, fourth and fifth since they appear to be an indefinite integral.

It sound picky, but you will come to realize the value when you do more advanced work.

Last edited: Oct 1, 2013
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8. ### DerStrom8 Well-Known Member

Feb 20, 2011
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Good catch there. I made that mistake once when designing something (can't remember what it was) and it gave me all sorts of trouble.

9. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21
Thank you very much. But how do you do that if the function being integrated has two or more terms? Will I just add the two limits in the integral sign and enclose the integrated function with brackets, with the two limits in the top and bottom of the closing bracket?

10. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
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The "+C" is put at the end, no matter how many terms you have. If you split up the terms, it's common to use a small (lower-case) "c" for each one. When you combine the parts, you use a big "+C"

11. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
It may or may not mean anything to you but

The definite integral is a pure number.

The indefinite integral is a function.

The constant C is arbitrary. This means that it can be any number we like.
Since it can be any number we can always add two or them together as DerStom said, to form one single constant.

Last edited: Oct 1, 2013
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12. ### WBahn Moderator

Mar 31, 2012
18,093
4,919
You need to be more careful about maintaining your notation. You are dropping things off and then carrying them mentally and applying them later from memory. That is asking for trouble.

$\int_2^3 2x^{3}+11x^{2}+31x dx$

While some people have no problem with grouping everything between the integral sign and the closing infinitesimal and calling it the integrand, it is generally better to use parens when you have multiple terms.

$\int_2^3 \left( 2x^{3}+11x^{2}+31x \right) dx$

When you break up the integral into multiple integrals, you need to carry along the limits (don't turn a definite integral into an indefinite one) and you can't just drop the infinitesimal.

$=\int_2^3 2x^{3} dx \, + \, \int_2^3 11x^{2} dx \, + \, \int_2^3 31x dx$

Technically (and this is a pretty minor point), since you chose to break your integral into three separate integrals, you've committed to evaluating them separately.

$= {\left\ \frac {x^4}{2} \right|_2^3} \, + \, {\left\ \frac{11x^3}{3} \right|_2^3} \, + \, {\left\ \frac{31x^2}{2} \right|_2^3}$

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13. ### Lightfire Thread Starter Well-Known Member

Oct 5, 2010
690
21

Hello,

What if we are asked to find the derivative of 5x.

We can do that by a simple formula:

$1(5x^{1-1})$

$=5x^{0}$

$=5$

Then we are now asked to find the integral of 5.

$\int 5dx$

$=\frac{5x^{0+1}}{1}$

$=5x^{1}+c$

$=5x+c$

But what if there is dx in the last part of the integrand?

Thank you very much.

14. ### WBahn Moderator

Mar 31, 2012
18,093
4,919
There is always an infinitesimal (e.g., dx) as the last part of the integrand.

Remember that the integral is essentially finding the area underneath a curve by adding up a series of small rectangular slices that are each f(x) high and Δx wide. The integral sign is basically a summation sign taken in the limit that Δx goes to zero and dx is basially Δx in that same limit.

15. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
Mathematicians look the other way now.

Lightfire you could look at it like this.

Say you wanted to integrate 3x. That means find the function whose derivative is 3x

$\frac{{dy}}{{dx}} = 3x$

$dy = 3xdx$

$\int {dy = \int {3xdx} }$

$y = \frac{3}{2}{x^2} + C$

Remember the constant?