Question

Integrals and area. Oh gosh I need help. Use area to evaluate the integral from 0 to b of (8x)/9 dx, b>0 ..I will draw a picture

Answer 1

8x/9 is just (8/9) x, so you can pull the constant 8/9 in front of the integral, and just integrate x dx.

Answer 2

\[\int\limits_{0}^{b} \frac{ 8x }{ 9 } dx , b >0\]

Answer 3

I'm sure you can integrate that.

Answer 4

Okay, I did that. Not sure what to do

Answer 5

What did you do. Let me see.

Answer 6

what's the general antiderivative of x?

Answer 7

would it be (8b)/9?

Answer 8

just leave the 8/9 out for now, and integrate xdx

Answer 9

What did you get for the integral of x dx?

Answer 10

@BreannaKawaii well?

Answer 11

(x^2)/2 ?

Answer 12

:D

Answer 13

yes.

Answer 14

my computer is broken so it's super slow and erases what I type halfway through typing

Answer 15

so what's\[\frac{x^2}{2} |_{0}^{b}\]

Answer 16

so evaluate x^2/2 from o to b; you get b^2/2 - 0 = b^2/2.

Now multiply b^2/2 by 8/9.

Answer 17

b^2/2

Answer 18

right, and we had an 8/9 out front

Answer 19

Ohhh okay..

Answer 20

multiply that by 8/9

Answer 21

Okay. I did

Answer 22

so what did you get?

Answer 23

8b^2/9

Answer 24

noooooo\[\frac{8}{9}\cdot \frac{b^2}{2}\]

Answer 25

you have b^2/2 multiplied by 8/9 should give you 8b^2/18 or 4b^2/9.

Answer 26

C'est vrai. Merci.

Answer 27

:D

Answer 28

welcome.