To solve the problem step by step, we need to calculate the integral \( \int_2^4 x g(x) \, dx \) given that \( \int g(x) \, dx = f^{-1}(x) \) and \( f(x) = x^3 + x + \sin(\pi x) + 2 \).
### Step 1: Verify that \( f(x) \) is an increasing function
To determine if \( f(x) \) is increasing, we differentiate \( f(x) \):
\[
f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x) + \frac{d}{dx}(\sin(\pi x)) + \frac{d}{dx}(2)
\]
Calculating each derivative:
- \( \frac{d}{dx}(x^3) = 3x^2 \)
- \( \frac{d}{dx}(x) = 1 \)
- \( \frac{d}{dx}(\sin(\pi x)) = \pi \cos(\pi x) \)
- \( \frac{d}{dx}(2) = 0 \)
Thus, we have:
\[
f'(x) = 3x^2 + 1 + \pi \cos(\pi x)
\]
Since \( 3x^2 \) and \( 1 \) are always positive, and \( \pi \cos(\pi x) \) oscillates but does not affect the positivity of the entire expression, we conclude that \( f'(x) > 0 \) for all \( x \). Therefore, \( f(x) \) is an increasing function.
### Step 2: Find \( f(0) \) and \( f(1) \)
We calculate:
\[
f(0) = 0^3 + 0 + \sin(0) + 2 = 2
\]
\[
f(1) = 1^3 + 1 + \sin(\pi) + 2 = 4
\]
### Step 3: Set up the integral \( \int_2^4 x g(x) \, dx \)
Using integration by parts, where \( u = x \) and \( dv = g(x) \, dx \):
\[
\int u \, dv = uv - \int v \, du
\]
Here, \( du = dx \) and \( v = f^{-1}(x) \) (since \( \int g(x) \, dx = f^{-1}(x) \)).
Thus, we have:
\[
\int_2^4 x g(x) \, dx = \left[ x f^{-1}(x) \right]_2^4 - \int_2^4 f^{-1}(x) \, dx
\]
### Step 4: Evaluate \( \left[ x f^{-1}(x) \right]_2^4 \)
Calculating the boundary terms:
1. For \( x = 4 \):
- Since \( f(1) = 4 \), we have \( f^{-1}(4) = 1 \).
- Thus, \( 4 f^{-1}(4) = 4 \times 1 = 4 \).
2. For \( x = 2 \):
- Since \( f(0) = 2 \), we have \( f^{-1}(2) = 0 \).
- Thus, \( 2 f^{-1}(2) = 2 \times 0 = 0 \).
Putting it together:
\[
\left[ x f^{-1}(x) \right]_2^4 = 4 - 0 = 4
\]
### Step 5: Calculate \( \int_2^4 f^{-1}(x) \, dx \)
We need to find \( \int_2^4 f^{-1}(x) \, dx \). To do this, we can use the area under the curve method:
1. The area under \( f(x) \) from \( 0 \) to \( 1 \) can be calculated as:
\[
\int_0^1 f(x) \, dx = \int_0^1 \left( x^3 + x + \sin(\pi x) + 2 \right) \, dx
\]
Calculating each term:
\[
\int_0^1 x^3 \, dx = \frac{1^4}{4} = \frac{1}{4}
\]
\[
\int_0^1 x \, dx = \frac{1^2}{2} = \frac{1}{2}
\]
\[
\int_0^1 \sin(\pi x) \, dx = -\frac{1}{\pi} \cos(\pi x) \bigg|_0^1 = -\frac{1}{\pi}(-1 - 1) = \frac{2}{\pi}
\]
\[
\int_0^1 2 \, dx = 2 \cdot 1 = 2
\]
Adding these results:
\[
\int_0^1 f(x) \, dx = \frac{1}{4} + \frac{1}{2} + \frac{2}{\pi} + 2 = \frac{1}{4} + \frac{2}{4} + \frac{2}{\pi} + 2 = \frac{3}{4} + \frac{2}{\pi} + 2
\]
### Step 6: Calculate the final integral
Now we can substitute back into our integration by parts result:
\[
\int_2^4 x g(x) \, dx = 4 - \left( \frac{3}{4} + \frac{2}{\pi} + 2 \right)
\]
Calculating:
\[
= 4 - \left( \frac{3}{4} + 2 + \frac{2}{\pi} \right) = 4 - \frac{11}{4} - \frac{2}{\pi} = \frac{16}{4} - \frac{11}{4} - \frac{2}{\pi} = \frac{5}{4} - \frac{2}{\pi}
\]
### Final Answer
Thus, the value of \( \int_2^4 x g(x) \, dx \) is:
\[
\frac{5}{4} + \frac{2}{\pi}
\]
