Question

The value of \(\int_{0}^{6} f(x)\,dx\), if \(f(x)=\begin{cases}x^2, & x < 2 \\ 3x-2, & x \ge 2\end{cases}\), is given by:

• A. 1
• B. 0
• C. $\frac{1}{3}$
• D. $\frac{128}{3}$

Hint:

For definite integrals of piecewise linear functions, you can also use basic geometry (like areas of trapezoids) to calculate the linear portion quickly:
Area of the trapezoid under $3x-2$ from $x=2$ to $x=6$: $\text{Height} = 4$, $\text{Parallel sides} = f(2)=4$ and $f(6)=16$.
$\text{Area} = \frac{1}{2} \times (4 + 16) \times 4 = 40$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To integrate a piecewise-defined function, we split the interval of integration at the boundary points where the function's definition changes.
Key Formula or Approach:
Using the additive property of definite integrals: \[ \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \]

Step 2: Detailed Explanation:
The given limits are from $0$ to $6$, and the function changes its definition at $x = 2$.
We partition the integral: \[ I = \int_{0}^{6} f(x) \, dx = \int_{0}^{2} x^2 \, dx + \int_{2}^{6} (3x-2) \, dx \] Evaluate the first integral: \[ \int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{8}{3} - 0 = \frac{8}{3} \] Evaluate the second integral: \[ \int_{2}^{6} (3x-2) \, dx = \left[ \frac{3x^2}{2} - 2x \right]_{2}^{6} \] At the upper limit $x = 6$: \[ \frac{3(6^2)}{2} - 2(6) = \frac{108}{2} - 12 = 54 - 12 = 42 \] At the lower limit $x = 2$: \[ \frac{3(2^2)}{2} - 2(2) = \frac{12}{2} - 4 = 6 - 4 = 2 \] Subtract the lower limit value from the upper limit value: \[ \int_{2}^{6} (3x-2) \, dx = 42 - 2 = 40 \] Combine both integrated parts: \[ I = \frac{8}{3} + 40 = \frac{8 + 120}{3} = \frac{128}{3} \]

Step 3: Final Answer:
The total value is $\frac{128}{3}$, which corresponds to Option (D).