Integral by interpreting it in terms of area

[tsukuba]

Homework Statement

∫(a= -3 , b= 0) (1 + √9 - x^2) dx

Homework Equations

∫(a,b) f(x) dx = lim as n → [itex]\infty[/itex] [itex]\sum[/itex] f(xi) delta x

The Attempt at a Solution

I tried plugging in my a and b value into the function just as I would with any other function to find the area and i get a number but the answer is a ∏ so I am not sure with the pi comes from

 

[jackarms]

What did you plug your numbers into? That's not an elementary integral. I think the intention was for you to think about what the graph of 1 + sqrt(9 - x^2) looks like, and then use geometry to calculate it.

 

[tsukuba]

I've been doing some research and i figure that 9-x^2 is 1/4 of a circle

 

[haruspex]

I've been doing some research and i figure that 9-x^2 is 1/4 of a circle

There is certainly a quarter circle involved, but the region is a bit more than that.

 

[tsukuba]

how do i know its a quarter circle though?

 

[jackarms]

A couple different ways. One is to rearrange the equation a bit.

Replace f(x) with y, and ignore the + 1 in front, and you have this:
$$y = \sqrt{9 - x^{2}}$$
With a little rearranging (square both sides, move x^{2} over), you get this:
$$x^{2} + y^{2} = 9$$
You can recognize that as the equation of a circle, with radius 3. So if plotted at the origin, it would stretch from -3 to 3 on both axes. Since the limits on integration are from x = -3 to x = 0, that's half of the circle. Then, since you take the positive square root, it's the half of the circle above the x-axis: thus, a quarter circle in the second quadrant. Look at the addition of 1 to the front of the original equation (f(x)), and you get that same quarter circle moved up one.