The value of \( \int_{-2}^{3}\displaystyle|x-1|\,dx \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 13/2
Since \( |x-1|=1-x \) for \( x\lt1 \) and \( x-1 \) for \( x\ge1 \), the integral splits at x = 1:
\( \displaystyle \int_{-2}^{1}(1-x)\,dx=\left[x-\frac{x^{2}}{2}\right]_{-2}^{1} \)
\( \displaystyle =\frac{1}{2}-(-4) \)
\( \displaystyle =\frac{9}{2} \)
\( \displaystyle \int_{1}^{3}(x-1)\,dx=\left[\frac{x^{2}}{2}-x\right]_{1}^{3} \)
\( \displaystyle =\frac{3}{2}-\left(-\frac{1}{2}\right) \)
\( =2 \)
Adding,
\( \displaystyle \frac{9}{2}+2=\frac{13}{2} \)
Shortcut: The integral is the area under the graph of \( |x-1| \), which is two triangles with vertex at x = 1:
\( \displaystyle \frac{1}{2}\times3\times3+\frac{1}{2}\times2\times2 \)
\( \displaystyle =\frac{13}{2} \)
Why the other options are incorrect
Option (1) 5/2 — Incorrect. 5/2 is \( \frac{9}{2}-2 \); the two areas must be added, not subtracted.
Option (2) −5/2 — Incorrect. −5/2 is what integrating \( x-1 \) over the whole interval gives, ignoring the modulus. An integral of \( |x-1| \) can never be negative.
Option (4) 13 — Incorrect. The value is 13/2, not 13.