To find the area enclosed by the curve \( y = x^2 + 3 \) and the lines \( x = 5 \) and \( y = 0 \), follow these steps:
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Determine the intersection points: We need to find the points where the curve intersects the line \( y = 0 \). Set the equation of the curve to zero:
\[ x^2 + 3 = 0 \]
This equation has no real solutions since \( x^2 + 3 \) is always positive (the minimum value is 3 when \( x = 0 \)). Thus, the curve does not intersect the line \( y = 0 \) within the real number system.
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Set up the area calculation: The area we want to calculate is bounded by the curve \( y = x^2 + 3 \), the vertical line \( x = 5 \), and the horizontal line \( y = 0 \). Since the curve is entirely above the x-axis, we can calculate the area from \( x = 0 \) (where the curve is at its lowest point) to \( x = 5 \).
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Calculate the area using integration: The area \( A \) between the curve and the x-axis from \( x = 0 \) to \( x = 5 \) can be found through the integral:
\[ A = \int_{0}^{5} (x^2 + 3) , dx \]
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Calculate the integral:
First, we compute the integral:
\[ A = \int_{0}^{5} (x^2 + 3) , dx = \int_{0}^{5} x^2 , dx + \int_{0}^{5} 3 , dx \]
We solve each part separately.
- For \( \int x^2 , dx \):
\[ \int x^2 , dx = \frac{x^3}{3} \Big|_{0}^{5} = \frac{5^3}{3} - \frac{0^3}{3} = \frac{125}{3} \]
- For \( \int 3 , dx \):
\[ \int 3 , dx = 3x \Big|_{0}^{5} = 3(5) - 3(0) = 15 \]
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Add the results:
Now, combine the two results:
\[ A = \frac{125}{3} + 15 = \frac{125}{3} + \frac{45}{3} = \frac{170}{3} \]
Thus, the exact area enclosed by the curve \( y = x^2 + 3 \), the line \( x = 5 \), and the line \( y = 0 \) is
\[ \boxed{\frac{170}{3}} \] square units.
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