What is the evaluated result of the integral ∫ from 1 to e of ln(t) dt?

What is the evaluated result of the integral ∫ from 1 to e of ln(t) dt?

• A. e - 1
• B. 1
• C. ln(e)
• D. 0

Answer: (B)

To evaluate the integral ∫ from 1 to e of ln(t) dt, we can use integration by parts. In integration by parts, we typically choose: - u = ln(t) which gives us du = (1/t) dt - dv = dt which results in v = t The integration by parts formula is given by ∫ u dv = uv - ∫ v du. Applying this to our integral, we have: ∫ ln(t) dt = t ln(t) - ∫ t (1/t) dt = t ln(t) - ∫ dt = t ln(t) - t + C, where C is the constant of integration. Now, we need to evaluate this expression at the bounds from 1 to e: 1. At the upper limit (t = e): e ln(e) - e = e(1) - e = e - e = 0. 2. At the lower limit (t = 1): 1 ln(1) - 1 = 1(0) - 1 = 0 - 1 = -1. Now, subtract the value at the lower limit from that at the upper limit: 0 - (-1

To evaluate the integral ∫ from 1 to e of ln(t) dt, we can use integration by parts. In integration by parts, we typically choose:

• u = ln(t) which gives us du = (1/t) dt

• dv = dt which results in v = t

The integration by parts formula is given by ∫ u dv = uv - ∫ v du. Applying this to our integral, we have:

∫ ln(t) dt = t ln(t) - ∫ t (1/t) dt

= t ln(t) - ∫ dt

= t ln(t) - t + C, where C is the constant of integration.

Now, we need to evaluate this expression at the bounds from 1 to e:

1. At the upper limit (t = e):

e ln(e) - e = e(1) - e = e - e = 0.

2. At the lower limit (t = 1):

1 ln(1) - 1 = 1(0) - 1 = 0 - 1 = -1.

Now, subtract the value at the lower limit from that at the upper limit:

0 - (-1