Question

Sketch a graph of \( y = x^2 \). Using integration, find the area of the region bounded by \( y = 9 \), \( x = 0 \), and \( y = x^2 \).

Hint:

When finding the area between curves, always find the points of intersection and use integration to subtract the lower curve from the upper curve.

Answer 1

The area between the curves \( y = x^2 \) and \( y = 9 \) is computed. The intersection points are found by setting \( x^2 = 9 \), yielding \( x = \pm 3 \). The integration interval is \( [-3, 3] \). The area \( A \) is the integral of the upper function (\( y = 9 \)) minus the lower function (\( y = x^2 \)) over this interval: \[ A = \int_{-3}^{3} (9 - x^2) \, dx \] This is evaluated as: \[ A = \int_{-3}^{3} 9 \, dx - \int_{-3}^{3} x^2 \, dx \] The first integral is \( \int_{-3}^{3} 9 \, dx = 9x \Big|_{-3}^{3} = 9(3) - 9(-3) = 54 \). The second integral is \( \int_{-3}^{3} x^2 \, dx = \frac{x^3}{3} \Big|_{-3}^{3} = \frac{3^3}{3} - \frac{(-3)^3}{3} = 9 - (-9) = 18 \). Therefore, the area is \( A = 54 - 18 = 36 \). The area is \( \boxed{36} \) square units.
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