# Help, integration, find areas

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• Jun 10th 2010, 05:03 AM
Tweety
Help, integration, find areas
Find the shaded areas,

I have attached the diagram, is this right set up for the area under the curve?

Also for the bit above the axis, the triangular bit, how would I work out this area? I dont know the equation of this line. Is it y = 4?

$\int^3_{-1} 0-x^{2} -2x -3\ dx$ this is the x-axis minus the curve.

Thank you!

Edit- just to clarify, the curve crosses the x-axis at -1 and 3, although on my diagram it looks like it doesn't
• Jun 10th 2010, 05:29 AM
sponsoredwalk
So the equation of the curve is;

$f(x) = x^2 - 2x - 3$

Well, when finding the area under a curve all you have to do is find the area under the top curve and then subtract the area from the bottom curve;
$
\int_{a}^{b} [f (x) - g(x)]\,dx
$

so, from -1 to 3 we see that the top curve is the x-axis and the bottom curve is f(x), then from 3 to 4 the top curve if f(x) and the bottom curve is y=0

$\int_{-1}^{3} [0 - (x^2 - 2x - 3)]\,dx + \int_{3}^{4} x^2 - 2x - 3 - (0)\,dx$

$\int_{-1}^{3} - x^2 + 2x + 3\,dx + \int_{3}^{4} x^2 - 2x - 3 \,dx$
• Jun 10th 2010, 05:43 AM
Tweety
Quote:

Originally Posted by sponsoredwalk
So the equation of the curve is;

$f(x) = x^2 - 2x - 3$

Well, when finding the area under a curve all you have to do is find the area under the top curve and then subtract the area from the bottom curve;
$
\int_{a}^{b} [f (x) - g(x)]\,dx
$

so, from -1 to 3 we see that the top curve is the x-axis and the bottom curve is f(x), then from 3 to 4 the top curve if f(x) and the bottom curve is y=0

$\int_{-1}^{3} [0 - (x^2 - 2x - 3)]\,dx + \int_{3}^{4} x^2 - 2x - 3 - (0)\,dx$

$\int_{-1}^{3} - x^2 + 2x + 3\,dx + \int_{3}^{4} x^2 - 2x - 3 \,dx$

thanks so much, makes sense, as I was trying to work out the height of the triangle, and not getting anywhere.

Also what is the equation of this straight line?

Just to clarify, could we have also worked out where this straight line intersects the curve, therefore the height, and use that to find the area or the triangle and add it to the area of the curve, found by integration?

Is this a valid method aswell?
• Jun 10th 2010, 06:09 AM
hungthinh92
Quote:

Originally Posted by Tweety
thanks so much, makes sense, as I was trying to work out the height of the triangle, and not getting anywhere.

Also what is the equation of this straight line?

Just to clarify, could we have also worked out where this straight line intersects the curve, therefore the height, and use that to find the area or the triangle and add it to the area of the curve, found by integration?

Is this a valid method aswell?

The area in the right of the picture is not a triangle, so you can't find the height or anything like that :D That area is made by a curve and a line.

I think the best way to solve this kind of problems is what sponsoredwalk showed to you