1. u-substitution

$\displaystyle \int (3x+1)^5\ dx \ \mbox{limits - lower} (-1), \mbox{upper -} \ (1 ) \mbox{answer I get is 2080/3 \ is this correct?}$

also - is there a way to write limits on the integral line?

2. Originally Posted by weezie23
$\displaystyle \int (3x+1)^5\ dx \ \mbox{limits - lower} (-1), \mbox{upper -} \ (1 ) \mbox{answer I get is 2080/3 \ is this correct?}$

also - is there a way to write limits on the integral line?
yes, type \int_{lower limit goes here}^{upper limit goes here} to get the limits on the integral sign

so you would type $$\int_{-1}^{1}(3x + 1)^5~dx$$ to get $\displaystyle \int_{-1}^{1}(3x + 1)^5~dx$

3. Originally Posted by weezie23
$\displaystyle \int (3x+1)^5\ dx \ \mbox{limits - lower} (-1), \mbox{upper -} \ (1 ) \mbox{answer I get is 2080/3 \ is this correct?}$

also - is there a way to write limits on the integral line?
To write subscript, use "_{script}", or, if the script is only one character long, "_s". To write superscript, use "^{script}", or "^s" for a single character. Please note that superscript and subscript can be written on top of each other, and in any place. For example, we have:

n_0^1

...yields...

$\displaystyle n_0^1$

So, for an integral, we just make the integral sign and then the limits...

\int_0^5dx

$\displaystyle \int_0^5dx$

For super/subscript longer than one character:

\int_{genesis}^{deuteronomy}inerrancy\;d(pentateuc h)=mosaic\;law

$\displaystyle \int_{genesis}^{deuteronomy}inerrancy\;d(pentateuc h)=mosaic\;law$

4. Originally Posted by weezie23
$\displaystyle \int (3x+1)^5\ dx \ \mbox{limits - lower} (-1), \mbox{upper -} \ (1 ) \mbox{answer I get is 2080/3 \ is this correct?}$

also - is there a way to write limits on the integral line?
$\displaystyle \displaystyle\int^{1}_{-1} (3x+1)^5 \ \mathrm{d}x$

Subsitute $\displaystyle u=3x+1$.

This integral works out nicely and you should get a whole number as the answer when you apply the limits.

5. Originally Posted by Jhevon

yes, type \int_{lower limit goes here}^{upper limit goes here} to get the limits on the integral sign

so you would type $$\int_{-1}^{1}(3x + 1)^5~dx$$ to get $\displaystyle \int_{-1}^{1}(3x + 1)^5~dx$

$\displaystyle \int_{-1}^{1}(3x + 1)^5~dx$ u=(3x+1) du=(3dx)

Lower limits x=-1 (3(-1)+1)=-2 Upper limits x=1 u=(3(1)+1)=4

$\displaystyle \frac{1}{3} \int_{-2}^{4} u^5 du$

I get 4,842,434

6. Originally Posted by weezie23
$\displaystyle \int_{-1}^{1}(3x + 1)^5~dx$ u=(3x+1) du=(3dx)

Lower limits x=-1 (3(-1)+1)=-2 Upper limits x=1 u=(3(1)+1)=4

$\displaystyle \frac{1}{3} \int_{-2}^{4} u^5 du$

I get 4,842,434
No, it's incorrect.

$\displaystyle \frac{1}{3} \int_{-2}^{4} u^5 \ \mathrm{d}u$

$\displaystyle =\frac13 \left[ \frac{u^6}{6}\right]_{-2}^{4}$