Question

definite integrals: problem 2
(1 point)
suppose
f(x)=\begin{cases}6 - 2x & \text{if } 0leq xlt3,\\2x - 6 & \text{if } 3leq xleq6.end{cases}
evaluate the definite integral by interpreting it in terms of signed area.
int_{0}^{6}f(x)dx=square

Explanation:

Step1: Analyze the first part of the function

For \(y = 6 - 2x\) (\(0\leq x<3\)):
When \(x = 0\), \(y=6\); when \(x = 3\), \(y=0\).
The area of the triangle formed by this line segment with the \(x\) - axis and \(y\) - axis is \(A_1=\frac{1}{2}\times3\times6 = 9\).

Step2: Analyze the second part of the function

For \(y = 2x - 6\) (\(3\leq x\leq6\)):
When \(x = 3\), \(y = 0\); when \(x = 6\), \(y=6\).
The area of the triangle formed by this line segment with the \(x\) - axis is \(A_2=\frac{1}{2}\times3\times6=9\).

Step3: Calculate the definite integral

Since the definite integral \(\int_{0}^{6}f(x)dx\) is the sum of the signed areas.
The function \(y = 6 - 2x\) (\(0\leq x<3\)) is above the \(x\) - axis (positive area) and \(y = 2x - 6\) (\(3\leq x\leq6\)) is also above the \(x\) - axis (positive area).
\(\int_{0}^{6}f(x)dx=A_1 + A_2\)

Answer:

\(18\)