## anonymous one year ago 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.

1. ali2x2

have you tried tigeralgebra.com?

2. ali2x2
3. OregonDuck

@ali2x2 are you going to help him?

4. ali2x2

@OregonDuck its ok you can help him

5. OregonDuck

nevermind i was going to suggest tiger too lol

6. Astrophysics

Your graph is |dw:1438787787036:dw| so your integral is $\int\limits_{a}^{b} [g(x)-f(x)]dx$ where your interval is a=0, b = 2

7. Astrophysics

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

8. Astrophysics

Now go ahead and integrate :-)

9. ali2x2

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

10. Astrophysics

|dw:1438788387136:dw| a clearer graph, where f(x) is red and g(x) is blue.

11. Astrophysics

So the area is |dw:1438788598493:dw| the idea for the integral is basically this $\int\limits_{a}^{b} [ \text{Top function-Bottom function}] dx$

12. Astrophysics

@Yaros Does that make sense?

13. Astrophysics

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