Problem 34 - Essential Calculus

Question

Find the area of the region bounded by the graphs of the equations. $$ y=-x^{2}+3 x, \quad y=0 $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs of the equations \(y=-x^{2}+3x\) and \(y=0\) is 4.5 square units.

Step 1: Find the x-intercepts of the function

The x-intercepts of the function \(y=-x^{2}+3x\) are the values of \(x\) for which \(y=0\). Hence, set \(y=-x^{2}+3x=0\). This equation can be factored as \(x(-x+3) = 0\). Solving for \(x\), we can get two solutions which are \(x=0\) and \(x=3\).

Step 2: Set up the integral

To find the area bounded by the graph of the function and the x-axis, we should integrate the function \(y=-x^{2}+3x\) from \(x=0\) to \(x=3\). This gives us the integral \(\int_{0}^{3} (-x^{2}+3x) dx\).

Step 3: Evaluate the integral

We have \(\int_{0}^{3} (-x^{2}+3x) dx = \left[-\frac{1}{3}x^{3}+\frac{3}{2}x^{2}\right]_{0}^{3}\), which simplifies to \(-\frac{1}{3}(3)^{3}+\frac{3}{2}(3)^{2}-\left(-\frac{1}{3}(0)^{3}+\frac{3}{2}(0)^{2}\right)= -9 + \frac{27}{2} = 4.5\).Thus, the area under the curve from x=0 to x=3 is 4.5 square units.

Step 4: Conclusion

Therefore, the area of the region bounded by the graphs of the equations \(y=-x^{2}+3x\) and \(y=0\) is 4.5 square units.