I just want a simple check to see where I made a mistake.

Question

Question) Sketch the graph on the interval -4 ≤ x ≤ 2 and calculate the definite integral in terms of signed area (do not anti differentiate).

f(x) = 4+x , for -4 ≤ x < -2
2-√(4-x^2) , for -2 ≤ x ≤ 2

I did the following work:

f(x) = 4 + x
x= -2, y = 4 - 2 = 2
x= -4, y= 4 - 4 = 0
This will be a straight line from -4 to -2

f(x) = 2-√(4-x^2)
x= -2, y = 2
x= 0, y = 0
x= 2, y = 2
This will be a upward parabola from -2 to 2

Here the graph of the function: desmos.com/calculator/yzjmlutrfy

Now the definite Integral from -4 to 2, where The first area is a triangle from [-4,-2], the second area is 1/4 a circle and the third area is also 1/4 a circle.

Definite Integral Area = A1 + A2 + A3
= (1/2 * 2 * 2) + (1/4 * pi * r^2) * (1/4 * pi * r^2)
= (4/2) + (4/4 * pi) + (4/4 * pi)
= 2 + pi + pi
= 2 + 2pi

My teacher answer = 10 - 2pi

Please help me figure out whats wrong. Thank you!

Answer 1

First off, √(4-x^2) is a semicircle (x^2+y^2 = 4), not a parabola
It is the upper semi-circle.
so, 2-√(4-x^2) is the lower semicircle, shifted up so it rests on the x-axis at (0,0)
The integral is just the area under the curves.

From -4 to -2 you just have a triangle of base 2 and height 2, so its area is 2.

Consider the 4x2 rectangle with vertices at (-2,2), (2,2), (2,0), (-2,0). Its area is 8
The area of the semi-circle is 2π, so the area under the curve is 8-2π
Thus, the total area (the value of the integral) is 10-2π