Evaluate the integral by interpreting it in terms of areas.

1) Evaluate the integral by interpreting it in terms of areas.
∫ (0 to -9) (5 + sqrt of (81 - x^2) ) dx

∫ (-5 + √(81-x^2)) dx from -9 to 0

∫ (-5) dx from -9 to 0 + ∫ √(81-x^2) dx from -9 to 0

The first integral represents the area above the horizontal line y = -5 from -9 to 0. Note that this is a rectangle of length 9 and width 5.

Area = -(9)(5) = -45

The second integral represents the area of a quarter circle of radius 9.

Area = π(9)^2/4 = 81π/4

Therefore the integral evaluates to 81π/4 - 45.

The i;ntegral of 5 dx = 5x] 0 to -9 = 45
The integral of (81 - x^2)dx We use the trignometric substitution
Draw a right angled trianle with an angle theta such that the side adjascent to theta is x and the hypontuse is 9, making the side opposite to theta (root 9^2 - x^2)
( 81-x^2 )^1/2 dx = (81 -x^2) /9 =Sin theta and (81- x^2) = 9 Sin theta
x/9 = Cos theta
x = 9 Cos theta
dx =- 9 Sin theta
= Intetral [ 9 Sin theta .( -9 Sjin theta) d theta= -Integral[ -81 Sin^2 theta) d theta
= -81[Intg ral 1/2 =1/2 Cos 2 theta d theta
= -81 [ 1/2 theta - 1/4 Sin 2 theta]
Limjits for theta: x = -9Cos theta
x=0 , theta pi/2
x = -9 . Theta = 0 = -1 [0- pi/2] - 81/4 [0 -0]
-
= - 127
Total area = 45 -127 - 82
I may have madae a mistake in the sign and m;aybe the answer should be positive.
Hope the procedure ils useful