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.

days of the week

step-by-step explanation:

javier earned more money then kyle, so that won't always be a constant term, but the days of the week do not change so that term is always constant.

edit: i meant