Thread: 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

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 .

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

Does that help ?

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

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.