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
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
Don't integrate - balloontegrate!
Balloon Calculus Forum
You are plugging the limits into your (correct) integral expression, are you?
$\displaystyle \int_4^6 (7x - 3x^3) dx = \large{[}\frac{7}{2}x^2 - \frac{3}{4}x^4\large{]}_4^6$
$\displaystyle = [\frac{7}{2}6^2 - \frac{3}{4}6^4] - [\frac{7}{2}4^2 - \frac{3}{4}4^4]$
= ...