Answer:
The area under the curve y = tsin(t-1), x = ㏑(t) from x = 0 to x = ㏑([tex]\pi[/tex] + 1) is 2 square units
Step-by-step explanation:
y = tsin(t-1)
x = ㏑(t)
⇒dx = [tex]\frac{1}{t}[/tex]dt
x = 0 ⇒ t = 1 and x = ㏑([tex]\pi[/tex]+1) ⇒ t = [tex]\pi[/tex]+1
Area under the curve from x = 0 to x = ㏑([tex]\pi[/tex]+1) or from t = 1 to t = [tex]\pi[/tex]+1 is [tex]\int\limits {y} \, dx[/tex] with limits : t from 1 to [tex]\pi[/tex]+1
= [tex]\int\limits {tsin(t-1)} \, \frac{1}{t} dt[/tex]
= [tex]\int\limit {sin(t-1)} \, dt[/tex]
Using the substitution t = k + 1 :
dt = dk
when t = 1, k = 0 and when t = [tex]\pi[/tex]+1, k = [tex]\pi[/tex]
The integral is now modified as :
[tex]\int\limits^\pi _0 {sin(k)} \, dk[/tex]
= -(cos([tex]\pi[/tex])-cos(0))
= -(-1-1)
= -(-2)
= 2 square units.
