How can I integrate the semicircle: integral from -2 to 2 (-v(4-x^2) +2)dx ? Thanks.

Question

Show steps. Thanks. The answer is 1.717, I believe. I checked using my calculator.

Answer 1

Hi Bob.

 

Here's where knowing the interpretation of the integral is more important than actually being able to calculate it.

 

An integral calculates the area underneath a curve. When finding the area underneath some curves it's easier to approach it geometrically than via calculus.

 

Recall from geometry that the area of a circle is pi*r², so a semi-circle must be 1/2*pi*r².

 

Take a look at your function. It looks like you're integrating the bottom half of a circle of radius 2 centered at (0,2).

 

Your region is going to look like a half-pipe. Break this region up into two shapes: larger rectangle, and a semi-circle being carved out of it. Find the area of the rectangle first, then subtract the area of the semi-circle inside it.

 

The large rectangle is going to have a width of 4 (since 2 - -2 = +4, which are your x bounds), and a height of 2, since that's where the circle is centered.

 

The area of the rectangular region will then be 4*2 = 8.

 

The area of your semi-circle will be:

 

1/2*pi*(2)² = 2pi

 

Since you want the half-pipe region below the semi-circle, your area will be:

 

Area of rectangle - Area of Semi-Circle

 

= 8 - 2pi

 

Which is about 1.717.

 

Hope this helps.