definte integrals

smartcar29 · #1 $Old$ July 4th, 2009, 06:00 AM

hi

can some one please check this for me

need to evaluate integral x(7-3x^2) for x=6 to x=4

i did this 1/2(7-3x^2)(2x)dx

1/2 *1/2(7-3x^2)^2+c

1/4(7-3x^2)^2+c

then for x=6 to x=4

1/4(7-3(6)^2-1/4(7-3(4)^2

am i on the right track ?

wayne

tom@ballooncalculus · #2 $Old$ July 4th, 2009, 06:18 AM

Some of your algebra is a bit worrying - check carefully.

Your integral is easy (no chain or product rule to work backwards through, just the power rule) if you start with

$x(7 - 3x^2) = 7x - 3x^3$

Then integrate each term separately.

tom@ballooncalculus · #3 $Old$ July 4th, 2009, 06:26 AM

Don't integrate - balloontegrate!

Balloon Calculus Forum

smartcar29 · #4 $Old$ July 4th, 2009, 06:34 AM

you came to my help again thanks

i get x(7-3x^2)dx to be

7x-3x^3
integrate= 7/2(x^2)-3/4(x^4)

if this is wrong then i think i will change my math course to history course.

tom@ballooncalculus · #5 $Old$ July 4th, 2009, 06:41 AM

That's fine, and brackets in order... hang on in there!

(Plug in the limits...)

smartcar29 · #6 $Old$ July 4th, 2009, 06:48 AM

plugged in limits and got answer to be 31.

thanks for you patients and help

wayne

tom@ballooncalculus · #7 $Old$ July 4th, 2009, 07:27 AM

You are plugging the limits into your (correct) integral expression, are you?

$\int_4^6 (7x - 3x^3) dx = \large{[}\frac{7}{2}x^2 - \frac{3}{4}x^4\large{]}_4^6$

$= [\frac{7}{2}6^2 - \frac{3}{4}6^4] - [\frac{7}{2}4^2 - \frac{3}{4}4^4]$

= ...

smartcar29 · #8 $Old$ July 4th, 2009, 07:46 AM

thanks for checking that now have -1094

tom@ballooncalculus · #9 $Old$ July 4th, 2009, 07:53 AM

Brackets! (Signs!)

smartcar29 · #10 $Old$ July 4th, 2009, 08:05 AM

eureka i have it -710

i get (7/2(6)^2)-(3/4(6)^4)= -846
and (7/2(4)^2)-(3/4(4)^4)= -136

-846--136= -710

if this is not correct i will eat my calculator.

tom@ballooncalculus · #11 $Old$ July 4th, 2009, 08:16 AM

Good. Then reflect that there are no negative areas...