Calculate area using intergral

[Loki123]

I seem to be having problems calculating the area of this triangle. My problem is that it's on the positive side of y axis and the negative. How do I define what I marked P2??

 

[pka]

Did you try [imath]\displaystyle\int_{ - \frac{5}{2}}^4 {\left( {[x + 1] - [\frac{{ - 3}}{2}]} \right)dx}~?[/imath]

 

[Loki123]

Did you try [imath]\displaystyle\int_{ - \frac{5}{2}}^4 {\left( {[x + 1] - [\frac{{ - 3}}{2}]} \right)dx}~?[/imath]

My teacher has been trying to explain that to me. But here is what I don't get:
He says to move the triangle up and calculate the area (P) like this:

But why can't I move it so it sits on the x-asis and calculate the area like this:

 

[pka]

To calculate the area between two curves using an integral is done by
approximating and adding areas of rectangles that fill the area between curves.
The base of the rectangle is [imath]dx[/imath] and the height is [imath]U(x)-L(x)[/imath] the upper minus the lower.

 

[BigBeachBanana]

My teacher has been trying to explain that to me. But here is what I don't get:
He says to move the triangle up and calculate the area (P) like this:
View attachment 31725
But why can't I move it so it sits on the x-asis and calculate the area like this:
View attachment 31729

The idea is to find the height of the triangle at every x-values (in blue). To do that we take [imath]f(x)-g(x)=(x+1)-\left(-\frac{3}{2}\right)[/imath] and integrate over all x-values in the domain.

The second idea or what you suggested is the same, except that you're finding the height of the triangle between f(x)=x+1 and g(x)=0.

 

[Loki123]

The idea is to find the height of the triangle at every x-values (in blue). To do that we take [imath]f(x)-g(x)=(x+1)-\left(-\frac{3}{2}\right)[/imath] and integrate over all x-values in the domain.

The second idea or what you suggested is the same, except that you're finding the height of the triangle between f(x)=x+1 and g(x)=0.

View attachment 31731

Got it, I think.

 

[Steven G]

You you like to can move this triangle so the bottom left point is at the origin--this seems to make you happy.
Just move all points to the right 5/2 units and up 3/2 units. Then integrate.

 

[Steven G]

Since you asked how to define P2 I will answer you.

\(\displaystyle P2 = \int_{-5/2}^{-1}(x+1)dx + \int_{-1}^4(0-(-3/2))dx\)

Of course to find p1, p2 or p1+p2 you do not need to use integrals