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

We are given the curves \( y = x^2 \) and \( y = 9 \). To find the area between these curves, we first determine the points of intersection. Set: \[ x^2 = 9 \] Solving this: \[ x = \pm 3 \] Thus, the curves intersect at \( x = -3 \) and \( x = 3 \). The area between the curves can be found by integrating the difference between the upper function (\( y = 9 \)) and the lower function (\( y = x^2 \)) over the interval \( [-3, 3] \). The area is: \[ A = \int_{-3}^{3} (9 - x^2) \, dx \] Breaking this into two integrals: \[ A = \int_{-3}^{3} 9 \, dx - \int_{-3}^{3} x^2 \, dx \] First, calculate the integral of \( 9 \): \[ \int_{-3}^{3} 9 \, dx = 9x \Big|_{-3}^{3} = 9(3) - 9(-3) = 9(6) = 54 \] Now, calculate the integral of \( x^2 \): \[ \int_{-3}^{3} x^2 \, dx = \frac{x^3}{3} \Big|_{-3}^{3} = \frac{(3)^3}{3} - \frac{(-3)^3}{3} = \frac{27}{3} - \frac{-27}{3} = 9 + 9 = 18 \] Subtracting the two integrals: \[ A = 54 - 18 = 36 \] Thus, the area of the region is \( \boxed{36} \) square units.
\begin{figure}[ht] \centering \includegraphics[width=0.7\textwidth]{ig4.png} \end{figure}