Academic...blah blah blah #3

Non-Calculator Portion Question 19

Question 19

The equation of the curve is y=(4)/(1+x^2). Find the area of the region formed by the line y=4 and the curve on the interval [-1,1].

1. Set up your definite integral.

(equation of top)-(equation of bottom)

!!!(can't really create a real integral problem using alt codes)!!!

(4-(4/(1+x^2)))

2. Integrate

(4-(4/(1+x^2)))

4x-...

4/(1+x^2) is apparently a arctan (or tan^(-1)) function, which is u'/(1+u^2).

4x-4(arctan(x)

3. Plug in your interval. Since the picture of the graph showed a symmetric region, you can use [0,1] as your a and b but just multiply your final answer by 2.

4(1)-4(arctan(1)) - [4(0)-4(arctan(0))]

4-4(π/4) - [0-4(0)]

4-π-0

2(4-π)=8-2π

8-2π is answer B and that is how you find it. It may look difficult but it is actually simple algebra.