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