Compute the definite integral:
integral_0^1 (5 x + 8)/(x^2 + 3 x + 2) dx
Rewrite the integrand (5 x + 8)/(x^2 + 3 x + 2) as (5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2)):
= integral_0^1 ((5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2))) dx
Integrate the sum term by term and factor out constants:
= 5/2 integral_0^1 (2 x + 3)/(x^2 + 3 x + 2) dx + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
For the integrand (2 x + 3)/(x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.
This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u): = (5 log(u))/2 right bracketing bar _2^6 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
Evaluate the antiderivative at the limits and subtract.
(5 log(u))/2 right bracketing bar _2^6 = (5 log(6))/2 - (5 log(2))/2 = (5 log(3))/2: = (5 log(3))/2 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx
For the integrand 1/(x^2 + 3 x + 2), complete the square:
= (5 log(3))/2 + 1/2 integral_0^1 1/((x + 3/2)^2 - 1/4) dx
For the integrand 1/((x + 3/2)^2 - 1/4), substitute s = x + 3/2 and ds = dx.
This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 1/(s^2 - 1/4) ds
Factor -1/4 from the denominator:
= (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 4/(4 s^2 - 1) ds
Factor out constants:
= (5 log(3))/2 + 2 integral_(3/2)^(5/2) 1/(4 s^2 - 1) ds
Factor -1 from the denominator:
= (5 log(3))/2 - 2 integral_(3/2)^(5/2) 1/(1 - 4 s^2) ds
For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds.
This gives a new lower bound p = (2 3)/2 = 3 and upper bound p = (2 5)/2 = 5:
= (5 log(3))/2 - integral_3^5 1/(1 - p^2) dp
Apply the fundamental theorem of calculus.
The antiderivative of 1/(1 - p^2) is tanh^(-1)(p):
= (5 log(3))/2 + (-tanh^(-1)(p)) right bracketing bar _3^5
Evaluate the antiderivative at the limits and subtract. (-tanh^(-1)(p)) right bracketing bar _3^5 = (-tanh^(-1)(5)) - (-tanh^(-1)(3)) = tanh^(-1)(3) - tanh^(-1)(5):
= (5 log(3))/2 + tanh^(-1)(3) - tanh^(-1)(5)
Which is equal to:
Answer: = log(18)
Given is :
The area of the rectangular patio is = 130 square feet
Let the width of the patio be = x feet
As length is 3 feet more than the width, so the length is = x+3 feet
Now area of rectangle is = length * width
This gives either x-10=0 or x+13=0
x-10=0 ; x=10
x+13=0 ; x=-13 (neglect this negative value)
So, the width of the patio is = 10 feet
Length = 10+3 = 13 feet
Let's begin by listing out the given information:
total number of members = 75
printing charge = $2 per shirt
colour print for each shirt = $50 fee for each color to be printed on the shirts
Let the number of t-shirts be represented as n
Let the number of colors on the shirts be represented as x
Let the total cost of the order be represented as C
Every member must have a t-shirt means
total number of members * printing charge + (colour print for each shirt * number of colors on the shirts) = total cost of the order
75 * 2 + 50 * x = C
150 + 50x = C
C = 50x + 150
The number of colors on the shirts (x) can vary change; if the number of colors used increases, the cost of the order increases & if the number decreases, the cost of the order decreases
The printing company charges is fixed as every member is to get a shirt
Answer:
6 spaces
Step-by-step explanation:
The vertical axis is on the left, and point B is moving to the right of that. Based on the number point B is aligned with on the horizontal axis, point B is 6 spaces away from the vertical axis.
Area of a triangle: 1/2 base x height
81 ft^2 = 1/2 (18 feet) (height)
Solve for x
81 / (1/2 x 18) = 9 feet for height
Please brainliest me if this helped; trying to level up!
