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This is the graph of the function f(x)=x2+4.Use integration to calculate the area of the shaded region. Give your answer without any rounding.

To calculate the area of the shaded region using integration, we need to first define the boundaries of the shaded region and then set up the integral of the given function.

### The function:
\[ f(x) = x^2 + 4 \]

### Boundaries:
From the graph, the shaded region appears to be bounded between \( x = -2 \) and \( x = -1 \).

### Area of the shaded region:
The area under the curve between these two bounds can be computed as:
\[
\text{Area} = \int_{-2}^{-1} (x^2 + 4) \, dx
\]

Now let's calculate this integral step-by-step.

1. The integral of \( x^2 \) is \( \frac{x^3}{3} \).
2. The integral of 4 is \( 4x \).

So, the integral becomes:
\[
\int_{-2}^{-1} (x^2 + 4) \, dx = \left[ \frac{x^3}{3} + 4x \right]_{-2}^{-1}
\]

Now let's compute this expression.The exact area of the shaded region is \( \frac{19}{3} \) square units.

Would you like further details or have any other questions? 

Here are some related questions:
1. How is the definite integral of a polynomial function computed in general?
2. How does changing the bounds affect the area calculation?
3. What happens if the function \( f(x) \) is shifted vertically or horizontally?
4. How can we visualize the meaning of the integral in terms of area?
5. What if the boundaries were set from \( x = 1 \) to \( x = 2 \)? How would the area change?

**Tip:** When calculating the area between the curve and the x-axis, always ensure that the function remains above the x-axis in the interval, otherwise you'll need to account for negative areas.

This is the graph of the function f(x) = x^2 + 4. Use integration to calculate the area of the shaded region. Give your answer without any rounding.

Learn how to calculate the area of the shaded region under the curve f(x) = x^2 + 4 between x = -2 and x = -1 using definite integrals. This problem covers key calculus concepts such as definite integrals and the fundamental theorem of calculus.

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