Question

find the area of the region bounded by the graphs of f(x) = x^3 + 4x^2 - 12x and g(x) = -x^2 +2x.
Please show the steps. No need to explaib.

Answer 1

have you tried tigeralgebra.com?

Answer 2

http://www.tiger-algebra.com/tiger.aspx

Answer 3

@ali2x2 are you going to help him?

Answer 4

@OregonDuck its ok you can help him

Answer 5

nevermind i was going to suggest tiger too lol

Answer 6

Your graph is so your integral is \[\int\limits_{a}^{b} [g(x)-f(x)]dx\] where your interval is a=0, b = 2

Answer 7

\[\int\limits_{0}^{2} [(-x^2+2x)-(x^3+4x^2-12x)] dx\]

Answer 8

Now go ahead and integrate :-)

Answer 9

umg i thought @OregonDuck already did the work, im missing out! ;o

Answer 10

a clearer graph, where f(x) is red and g(x) is blue.

Answer 11

So the area is the idea for the integral is basically this \[\int\limits_{a}^{b} [ \text{Top function-Bottom function}] dx\]

Answer 12

@Yaros Does that make sense?

Answer 13

Hey, so just to make sure you got it, \[\int\limits\limits\limits_{0}^{2} [(-x^2+2x)-(x^3+4x^2-12x)] dx = \int\limits\limits_{0}^{2} (-x^3-5x^2+14x)dx\]
\[[\frac{ -x^4 }{ 4 }-\frac{ 5x^3 }{ 3 }+7x^2] |_{0}^2 =-\frac{ (2)^4 }{ 4 }-\frac{ 5(2)^3 }{ 3 }+7(2)^2-0 = \frac{ 32 }{ 3 }\]