# Thread: Area of the region?

1. ## Area of the region?

guuh, I've been working on this problem forever and can't get it.

Find the area of the region bounded by the graphs f(x) = -x^2 + 7x + 20 and g(x) = x^2 + 5x - 40

To get the left and right x values

I do f(x) = g(x)

x^2 + 7x + 20 = x^2 + 5x - 40

2x^2 - 2x - 60 = 0

but then I'm stuck

... I think i know what to do once I get the values though

2. Originally Posted by Calculusretard
guuh, I've been working on this problem forever and can't get it.

Find the area of the region bounded by the graphs f(x) = -x^2 + 7x + 20 and g(x) = x^2 + 5x - 40

To get the left and right x values

I do f(x) = g(x)

x^2 + 7x + 20 = x^2 + 5x - 40

2x^2 - 2x - 60 = 0

but then I'm stuck

... I think i know what to do once I get the values though
Since you're studying calculus you're expected to be able to solve a quadratic equation. The x-coordinates of the intersection points are x = 6 and x = -5.

Draw a graph of the two curves and shade the required region. You should know that the enclosed area is given by $\displaystyle \int_{-5}^{6} \text{Upper curve} - \text{Lower curve} \, dx$.

It should be crystal clear from your graphs which is the upper curve and which is the lower curve. The integration is routine.

3. $\displaystyle 2x^{2} -2x -60 =0$

$\displaystyle x^{2} -x -30 = 0$

$\displaystyle (x-6)(x+5)=0$

$\displaystyle x = -5 ,6$

Does that help ?

4. Ick. I got the wrong answer >_<

Thank you both for the responses!! Can either of you please tell me what I'm doing wrong/how to do it right!?

6
∫ -x^2 + 7x +20 - (x^2 +5x - 40) dx
-5

6
∫ -2x^2 + 2x + 60
-5

(-2x^3)/ 3 + (2x^2)/ 2

(-2(6)^3)/ 3 + (2(6)^2)/ 2 - ((-2(-5)^3)/ 3 + (2(-5)^2)/ 2)

And then I simplified it until I got 865/3 ... which is the wrong answer

5. Originally Posted by Calculusretard
Ick. I got the wrong answer >_<

Thank you both for the responses!! Can either of you please tell me what I'm doing wrong/how to do it right!?

6
∫ -x^2 + 7x +20 - (x^2 +5x - 40) dx
-5

6
∫ -2x^2 + 2x + 60
-5

(-2x^3)/ 3 + (2x^2)/ 2

(-2(6)^3)/ 3 + (2(6)^2)/ 2 - ((-2(-5)^3)/ 3 + (2(-5)^2)/ 2)

And then I simplified it until I got 865/3 ... which is the wrong answer
The graph of the region.