Using property $\mathbf{P_2}$, $\displaystyle\int_{-1}^{2} |x^3-x|\,dx$ equals
A$\dfrac{11}{4}$
B$\dfrac{3}{2}$
C$2$
D$\dfrac{9}{4}$
Answer & Solution
Correct answer: A. $\dfrac{11}{4}$
1. $x^3-x\ge 0$ on $[-1,0]$ and $[1,2]$, but $\le 0$ on $[0,1]$.
2. Split: $\int_{-1}^0(x^3-x)+\int_0^1(x-x^3)+\int_1^2(x^3-x)$.
3. Each piece uses $\dfrac{x^4}{4}-\dfrac{x^2}{2}$; evaluating gives $\tfrac14+\tfrac14+\tfrac94$.
4. Sum $=\dfrac{11}{4}$. Dropping the sign change over $[0,1]$ gives the wrong value.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.343_
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