Questions: Determine the area of the given region. y=x+sin(x)

Solution Steps

The given region is bounded by the curve \( y = x + \sin(x) \), the x-axis, and the vertical lines \( x = 0 \) and \( x = \pi \).

To find the area under the curve from \( x = 0 \) to \( x = \pi \), we set up the definite integral: \[ \text{Area} = \int_{0}^{\pi} (x + \sin(x)) \, dx \]

We integrate the function \( x + \sin(x) \): \[ \int (x + \sin(x)) \, dx = \int x \, dx + \int \sin(x) \, dx \]

For the first part: \[ \int x \, dx = \frac{x^2}{2} \]

For the second part: \[ \int \sin(x) \, dx = -\cos(x) \]

Combining these results: \[ \int (x + \sin(x)) \, dx = \frac{x^2}{2} - \cos(x) + C \]

Evaluate the integral from \( x = 0 \) to \( x = \pi \): \[ \left[ \frac{x^2}{2} - \cos(x) \right]_{0}^{\pi} \]

Substitute the upper limit \( x = \pi \): \[ \left( \frac{\pi^2}{2} - \cos(\pi) \right) = \frac{\pi^2}{2} - (-1) = \frac{\pi^2}{2} + 1 \]

Substitute the lower limit \( x = 0 \): \[ \left( \frac{0^2}{2} - \cos(0) \right) = 0 - 1 = -1 \]

Subtract the lower limit result from the upper limit result: \[ \left( \frac{\pi^2}{2} + 1 \right) - (-1) = \frac{\pi^2}{2} + 1 + 1 = \frac{\pi^2}{2} + 2 \]

Final Answer