Let 9x^2-1 = y^2
<span>=> 18xdx = 2ydy </span>
<span>=> ydy = 9xdx </span>
<span>lower limit = sqrt(9*2/9 - 1) = sqrt(1) = 1 </span>
<span>upper limit = sqrt(9*4/9 - 1) = sqrt(3) </span>
<span>Int. [sqrt(2)/3,2/3] 1/(x^5(sqrt(9x^2-1)) dx </span>
<span>= Int. [sqrt(2)/3,2/3] xdx/(x^6(sqrt(9x^2-1)) </span>
<span>= 81* Int. [1,sqrt(3)] ydy/((y^2+1)^3y) </span>
<span>=81* Int. [1,sqrt(3)] dy/(y^2+1)^3 </span>
<span>y=tanz </span>
<span>dy = sec^2z dz </span>
<span>=81*Int [pi/4,pi/3] cos^4(z) dz </span>
<span>=81/4*int [pi/4,pi/3] (1+cos(2z))^2 dz </span>
<span>=81/4* Int. [pi/4,pi/3] (1+2cos(2z)+cos^2(2z)) dz </span>
<span>=81/4*(pi/3-pi/4) + 81/4*(sin(2pi/3)-sin(pi/2)) + 81/8 * (pi/3-pi/4) </span>
<span>+ 81/32 *(sin(-pi/3)-sin(pi)) </span>
<span>=81(pi/48+pi/96+1/4*(sqrt(3)/2 - 1) - 1/32 * sqrt(3)/2) </span>
<span>=81/32*(pi+3sqrt(3)-8)</span>
Answer:
$7.50
Step-by-step explanation:
75% of $10
=3/4×$10
=7.5
Area is all the space that is taken up by a two dimensional object. (the inside)
Perimeter is the distance around a two dimensional object. (the outside)
Answer:
0.3 with the hat ? the answer is first no, then it's a repeated decimal.
Step-by-step explanation:
