The value of the integral \( \int_{-6}^{0}\displaystyle|x+3|\,dx \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9
\( x+3 \) changes sign at \( x=-3 \), which lies between the limits, so the integral splits there:
\( \displaystyle |x+3|=-(x+3)\ \text{for}\ -6\le x\le -3 \)
\( \displaystyle |x+3|=x+3\ \text{for}\ -3\le x\le 0 \)
\( \displaystyle \int_{-6}^{0}|x+3|\,dx=\int_{-6}^{-3}-(x+3)\,dx+\int_{-3}^{0}(x+3)\,dx \)
\( \displaystyle \int_{-6}^{-3}-(x+3)\,dx=-\left[\frac{x^2}{2}+3x\right]_{-6}^{-3} \)
\( \displaystyle =-\left[\left(\frac{9}{2}-9\right)-(18-18)\right] \)
\( \displaystyle =\frac{9}{2} \)
\( \displaystyle \int_{-3}^{0}(x+3)\,dx=\left[\frac{x^2}{2}+3x\right]_{-3}^{0} \)
\( \displaystyle =0-\left(\frac{9}{2}-9\right) \)
\( \displaystyle =\frac{9}{2} \)
\( \displaystyle \text{Total}=\frac{9}{2}+\frac{9}{2}=9 \)
Shortcut: The graph of \( y=|x+3| \) on [−6, 0] is a V with its vertex at (−3, 0). The area under it is two triangles, each with base 3 and height 3:
\( \displaystyle 2\times\frac{1}{2}\times 3\times 3=9 \)
Why the other options are incorrect
Option (1) 9/4 — Incorrect. This is half of one triangle; each triangle has area 9/2.
Option (2) 1/2 — Incorrect. The value is 9, not 1/2.
Option (4) 9/2 — Incorrect. This is only one of the two equal pieces; the part from −6 to −3 must be added.