Question

Find the area of the region enclosed between the curve y=12x^2 + 30x and the x-axis.

Answer 1

i got this, doing the same thing fam

Answer 2

first you find the point of intersection

Answer 3

My answer to the problem was -625/4 but the book says its 125/4.

Answer 4

maybe your top curve was wrong

Answer 5

Let me type down from where I think I made an error.

Answer 6

\[\int\limits_{0}^{-5/2} 12x ^{2}+30x dx\]

Answer 7

\[\left[ 4x ^{3}+15x ^{2} \right]_{0}^{-5/2} +k\]

Answer 8

wrong way

Answer 9

0 on the top and -5/2 on bottom

Answer 10

OK, but I still end up with the same answer because when I evaluate the definite integral I'm still going to get a negative answer.

Answer 11

i got positive

Answer 12

125/4

Answer 13

Can you show your work plox

Answer 14

12x^2+30x= 0
so the intersection points are x = 0 and -5/2

Answer 15

rite

Answer 16

top curve is 12x^2+30x because you evaluate a number between 0 and -5/2 and it is greater than 0

Answer 17

wait

Answer 18

same

Answer 19

yea its negative number

Answer 20

-156.25 .....

Answer 21

I know to make it positive I need to swap a subtraction sign for and addition sign but I don't the proper procedure.

Answer 22

\[12x^2+30x=0 \\ x=0 \text{ or } 12x+30=0 \\ x=0 \text{ or } x=\frac{-30}{12} \\ x=0 \text{ or } x=\frac{-5}{2} \]
your limits are right
upper limit=0
lower limit=-5/2

now we need to figure out which function is greater y=12x^2+30x or y=0

well you could plug in a number between the intersections to see.(Or you look at graph)
I will do it the non-calculator way.
A number between -5/2 and 0 is -1.
y(x)=12x^2+30x
y(-1)=12(-1)^2+30(-1)=12-30=-18
and
y(x)=0
y(0)=0

so since -18<0 then 12x^2+30x<0 on the interval (-5/2,0)
so y=0 is the greater function (The one above the other curve)

\[\int\limits_\frac{-5}{2}^0 (0-[12x^2+30x]) dx\]

Answer 23

\[\int\limits_\frac{-5}{2}^0 (-12x^2-30x) dx\]

Answer 24

this will give you a positive result (as area should be positive or zero)

Answer 25

Ah-ha! Thank you so much!

Answer 26

np