Find the area under the curve y = tsin(t-1), x = ln(t) from x = 0 to x = ln(pi + 1)

The area under the curve y = tsin(t-1), x = ㏑(t) from x = 0 to x = ㏑( + 1) is 2 square units

Step-by-step explanation:

y = tsin(t-1)

x = ㏑(t)

⇒dx = dt

x = 0 ⇒ t = 1 and x = ㏑(+1) ⇒ t = +1

Area under the curve from x = 0 to x = ㏑(+1) or from t = 1 to t = +1 is with limits : t from 1 to +1

                       =

                       =

Using the substitution t = k + 1 :

dt = dk

when t = 1, k = 0 and when t = +1, k =

The integral is now modified as :

                 = -(cos()-cos(0))

                 = -(-1-1)

                 = -(-2)

                 = 2 square units.

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

22 and 22

step-by-step explanation:

10 plus 12 equals 22.

12 plus 10 also equals 22 because regardless of what order they're in, they're gonna equal the same thing.