Question

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

Answer 1

Answer:

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

Step-by-step explanation:

y = tsin(t-1)

x = ㏑(t)

⇒dx = $\frac{1}{t}$dt

x = 0 ⇒ t = 1 and x = ㏑($\pi$+1) ⇒ t = $\pi$+1

Area under the curve from x = 0 to x = ㏑($\pi$+1) or from t = 1 to t = $\pi$+1 is $\int\limits {y} \, dx$ with limits : t from 1 to $\pi$+1

                       = $\int\limits {tsin(t-1)} \, \frac{1}{t} dt$

                       = $\int\limit {sin(t-1)} \, dt$

Using the substitution t = k + 1 :

dt = dk

when t = 1, k = 0 and when t = $\pi$+1, k = $\pi$

The integral is now modified as :

$\int\limits^\pi _0 {sin(k)} \, dk$

                 = -(cos($\pi$)-cos(0))

                 = -(-1-1)

                 = -(-2)

                 = 2 square units.