\( \int_{0}^{2}\displaystyle|x^2-1|dx= \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
\( x^2-1 \) changes sign at x = 1 in [0, 2]:
\( \displaystyle 0\le x\le 1:\ |x^2-1|=1-x^2 \)
\( \displaystyle 1\le x\le 2:\ |x^2-1|=x^2-1 \)
\( \displaystyle \int_{0}^{2}\displaystyle|x^2-1|\,dx \)
\( \displaystyle =\int_{0}^{1}\displaystyle(1-x^2)\,dx+\int_{1}^{2}\displaystyle(x^2-1)\,dx \)
\( \displaystyle =\left[x-\frac{x^3}{3}\right]_{0}^{1}+\left[\frac{x^3}{3}-x\right]_{1}^{2} \)
\( \displaystyle =\frac{2}{3}+\left[\left(\frac{8}{3}-2\right)-\left(\frac{1}{3}-1\right)\right] \)
\( \displaystyle =\frac{2}{3}+\frac{4}{3}=2 \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{8}{3} \) — Incorrect. \( \frac{8}{3} \) is \( \int_{0}^{2}x^2\,dx \); the −1 and the modulus are both missing.
Option (3) \( \displaystyle\frac{14}{3} \) — Incorrect. \( \frac{14}{3} \) is \( \int_{0}^{2}(x^2+1)\,dx \), which is not the given integrand.
Option (4) \( \displaystyle\frac{2}{3} \) — Incorrect. \( \frac{2}{3} \) is \( \int_{0}^{2}(x^2-1)\,dx \), without the modulus: the negative part on [0, 1] cancels part of the positive part instead of being added.