Question

If \(f(x)=\begin{cases} 2x+8, & 1\le x\le2 \\ 6x, & 2\lt x \lt4\end{cases}\), then \(\int_1^4f(x)\) is :

• A. 43
• B. 45
• C. 47
• D. 46

Answer 1

The Correct Option is C

To find the definite integral of the piecewise function \( f(x) \) from 1 to 4, we need to evaluate each piece separately over its respective interval and then sum the results.

Given: \( f(x) = \begin{cases} 2x + 8, & 1 \le x \le 2 \\ 6x, & 2 < x \le 4 \end{cases} \)

We break down the integral \(\int_1^4 f(x) \, dx \) into two parts: \(\int_1^2 (2x+8) \, dx\) and \(\int_2^4 6x \, dx\).

1. For the first interval \(1 \le x \le 2\), compute \(\int_1^2 (2x+8) \, dx\):

- The integral of \((2x+8)\) is: \(\int (2x+8) \, dx = x^2 + 8x + C\).

- Evaluate from 1 to 2:

\[ \left[ x^2 + 8x \right]_1^2 = (2)^2 + 8(2) - \left[(1)^2 + 8(1)\right] = (4 + 16) - (1 + 8) = 20 - 9 = 11 \]

2. For the second interval \(2 < x \le 4\), compute \(\int_2^4 6x \, dx\):

- The integral of \(6x\) is: \(\int 6x \, dx = 3x^2 + C\).

- Evaluate from 2 to 4:

\[ \left[ 3x^2 \right]_2^4 = 3(4^2) - 3(2^2) = 48 - 12 = 36 \]

3. Sum the evaluated values from both intervals: \(11 + 36 = 47\)

Thus, the value of \(\int_1^4 f(x) \, dx\) is:

\[ \boxed{47} \]