Question

The area enclosed by $C_1$: y = Inx and $C_2$: y = $tan^{-1}$(x - 1) between the vertical lines x = 1 and x = 2 (in square units) is equal to

Answer 1

$1 + \frac{\pi}{4} - \frac{5}{2} \ln 2$

Answer 2

To find the area enclosed by the curves $C_1$: y = ln x and $C_2$: y = tan⁻¹(x - 1) between x = 1 and x = 2, we first need to determine which curve is above the other in the given interval.

1. Analyze the functions at the boundaries:

   • At x = 1:
     • For $C_1$: y = ln(1) = 0.
     • For $C_2$: y = tan⁻¹(1 - 1) = tan⁻¹(0) = 0. Both curves pass through the point (1, 0).
   • At x = 2:
     • For $C_1$: y = ln(2) ≈ 0.693.
     • For $C_2$: y = tan⁻¹(2 - 1) = tan⁻¹(1) = π/4 ≈ 0.785. Since π/4 > ln(2), at x = 2, $C_2$ is above $C_1$.

2. Determine which function is greater in the interval (1, 2): Let's consider the function f(x) = tan⁻¹(x - 1) - ln x. We want to find the sign of f(x) for x ∈ (1, 2). We know f(1) = 0. Let's find the derivative f'(x): $f'(x) = \frac{d}{dx}(\tan^{-1}(x-1)) - \frac{d}{dx}(\ln x)$ $f'(x) = \frac{1}{1+(x-1)^2} - \frac{1}{x}$ $f'(x) = \frac{x - (1+(x-1)^2)}{x(1+(x-1)^2)}$ $f'(x) = \frac{x - (1+x^2-2x+1)}{x(1+(x-1)^2)}$ $f'(x) = \frac{x - x^2 + 2x - 2}{x(1+(x-1)^2)}$ $f'(x) = \frac{-x^2 + 3x - 2}{x(1+(x-1)^2)}$ $f'(x) = \frac{-(x^2 - 3x + 2)}{x(1+(x-1)^2)}$ $f'(x) = \frac{-(x-1)(x-2)}{x(1+(x-1)^2)}$ For x ∈ (1, 2):

   • (x - 1) > 0
   • (x - 2) < 0
   • So, (x - 1)(x - 2) < 0.
   • Therefore, -(x - 1)(x - 2) > 0.
   • The denominator x(1 + (x - 1)²) is positive for x ∈ (1, 2). Thus, f'(x) > 0 for x ∈ (1, 2). Since f(1) = 0 and f(x) is increasing in (1, 2), it implies f(x) > 0 for x ∈ (1, 2). Therefore, tan⁻¹(x - 1) > ln x for x ∈ (1, 2). The area is given by the integral of the upper curve minus the lower curve: Area = $\int_{1}^{2} (\tan^{-1}(x-1) - \ln x) dx$ Area = $\int_{1}^{2} \tan^{-1}(x-1) dx - \int_{1}^{2} \ln x dx$

3. Evaluate $\int \tan^{-1}(x-1) dx$: Let u = x - 1, so du = dx. The integral becomes $\int \tan^{-1}(u) du$. Using integration by parts, $\int v dw = vw - \int w dv$: Let v = tan⁻¹(u) and dw = du. Then dv = $\frac{1}{1+u^2} du$ and w = u. $\int \tan^{-1}(u) du = u \tan^{-1}(u) - \int \frac{u}{1+u^2} du$ For the second integral, let t = 1 + u², so dt = 2u du, which means u du = $\frac{1}{2} dt$. $\int \frac{u}{1+u^2} du = \int \frac{1}{2t} dt = \frac{1}{2} \ln|t| = \frac{1}{2} \ln(1+u^2)$. So, $\int \tan^{-1}(u) du = u \tan^{-1}(u) - \frac{1}{2} \ln(1+u^2)$. Substituting back u = x - 1: $\int \tan^{-1}(x-1) dx = (x-1) \tan^{-1}(x-1) - \frac{1}{2} \ln(1+(x-1)^2)$. Evaluating the definite integral from 1 to 2: $[ (2-1)\tan^{-1}(2-1) - \frac{1}{2} \ln(1+(2-1)^2) ] - [ (1-1)\tan^{-1}(1-1) - \frac{1}{2} \ln(1+(1-1)^2) ]$ $= [ 1 \cdot \tan^{-1}(1) - \frac{1}{2} \ln(1+1) ] - [ 0 \cdot \tan^{-1}(0) - \frac{1}{2} \ln(1+0) ]$ $= [ \frac{\pi}{4} - \frac{1}{2} \ln 2 ] - [ 0 - 0 ]$ $= \frac{\pi}{4} - \frac{1}{2} \ln 2$.

4. Evaluate $\int \ln x dx$: Using integration by parts: Let v = ln x and dw = dx. Then dv = $\frac{1}{x} dx$ and w = x. $\int \ln x dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - \int 1 dx = x \ln x - x$. Evaluating the definite integral from 1 to 2: $[ 2 \ln 2 - 2 ] - [ 1 \ln 1 - 1 ]$ $= [ 2 \ln 2 - 2 ] - [ 0 - 1 ]$ $= 2 \ln 2 - 2 + 1$ $= 2 \ln 2 - 1$.

5. Calculate the total area: Area = $(\frac{\pi}{4} - \frac{1}{2} \ln 2) - (2 \ln 2 - 1)$ Area = $\frac{\pi}{4} - \frac{1}{2} \ln 2 - 2 \ln 2 + 1$ Area = $\frac{\pi}{4} + 1 - (\frac{1}{2} + 2) \ln 2$ Area = $\frac{\pi}{4} + 1 - (\frac{1+4}{2}) \ln 2$ Area = $1 + \frac{\pi}{4} - \frac{5}{2} \ln 2$.

The final answer is $\boxed{1 + \frac{\pi}{4} - \frac{5}{2} \ln 2}$.