If \(f(x) = \begin{cases} 2x + 8, & 1 \le x \le 2 \\ 6x, & 2 \lt x \lt 4 \end{cases}\), then \(\int_{1}^{4}\displaystyle f(x)\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 47
f changes its formula at x = 2, so the integral is split at x = 2:
\( \displaystyle \int_{1}^{4} f(x)\,dx = \int_{1}^{2} (2x + 8)\,dx + \int_{2}^{4} 6x\,dx \)
First part:
\( \displaystyle \int_{1}^{2} (2x + 8)\,dx = \Big[x^2 + 8x\Big]_{1}^{2} \)
\(= (4 + 16) - (1 + 8) = 11\)
Second part:
\( \displaystyle \int_{2}^{4} 6x\,dx = \Big[3x^2\Big]_{2}^{4} \)
\(= 48 - 12 = 36\)
Adding:
\(11 + 36 = 47\)