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

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