1. ## definte integrals

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

2. 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.

3. Don't integrate - balloontegrate!

Balloon Calculus Forum

4. 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.

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

(Plug in the limits...)

6. plugged in limits and got answer to be 31.

thanks for you patients and help

wayne

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

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

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

= ...

8. thanks for checking that now have -1094

9. Brackets! (Signs!)

10. 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.

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