Question

find the area of the region between the curves y = x(3 - x) and y = 3 from x = 0 to x = 3. the area is \boxed{} square units. (simplify your answer. type an integer or a fraction.)

Explanation:

Step1: Determine the upper and lower functions

First, we need to see which function is on top between \( y = 3 \) and \( y=x(3 - x)=3x - x^{2}\) from \( x = 0 \) to \( x = 3 \). Let's pick a test point, say \( x = 1 \). For \( y = 3x - x^{2}\), when \( x = 1 \), \( y=3(1)-1^{2}=2\), and \( y = 3 \) is greater than 2. So \( y = 3 \) is the upper function and \( y=3x - x^{2}\) is the lower function.

Step2: Set up the integral for the area

The formula for the area between two curves \( y = f(x) \) (upper) and \( y = g(x) \) (lower) from \( x = a \) to \( x = b \) is \( A=\int_{a}^{b}[f(x)-g(x)]dx \). Here, \( f(x)=3 \), \( g(x)=3x - x^{2}\), \( a = 0 \), and \( b = 3 \). So the area \( A=\int_{0}^{3}[3-(3x - x^{2})]dx=\int_{0}^{3}(3 - 3x+x^{2})dx \).

Step3: Integrate the function

We integrate term by term. The integral of 3 with respect to \( x \) is \( 3x \), the integral of \( - 3x \) is \( -\frac{3}{2}x^{2}\), and the integral of \( x^{2} \) is \( \frac{1}{3}x^{3} \). So the antiderivative \( F(x)=3x-\frac{3}{2}x^{2}+\frac{1}{3}x^{3} \).

Step4: Evaluate the definite integral

We evaluate \( F(x) \) from 0 to 3. \( F(3)=3(3)-\frac{3}{2}(3)^{2}+\frac{1}{3}(3)^{3}=9-\frac{27}{2}+9 \). \( F(0)=3(0)-\frac{3}{2}(0)^{2}+\frac{1}{3}(0)^{3}=0 \). Then \( A = F(3)-F(0)=9-\frac{27}{2}+9-0 \). Combine like terms: \( 18-\frac{27}{2}=\frac{36 - 27}{2}=\frac{9}{2} \).

Answer:

$\frac{9}{2}$