Home › ISC Class 12 › Mathematics › Application of Integrals › The value of $\int_{-6}^{0} |x + 3|\,dx$ is
The value of $\int_{-6}^{0} |x + 3|\,dx$ is
A$9$
B$0$
C$18$
D$6$
Answer & Solution
Correct answer: A. $9$
1. $|x + 3|$ has a corner at $x = -3$; split there: $|x+3| = -(x+3)$ on $[-6, -3]$ and $(x+3)$ on $[-3, 0]$.
2. $\int_{-6}^{-3}-(x+3)\,dx = \left[-\frac{x^2}{2} - 3x\right]_{-6}^{-3} = \frac{9}{2}$.
3. $\int_{-3}^{0}(x+3)\,dx = \left[\frac{x^2}{2} + 3x\right]_{-3}^{0} = \frac{9}{2}$.
4. Adding, the value $= \frac{9}{2} + \frac{9}{2} = 9$.
5. Each piece is a right triangle of legs $3$ and $3$ (area $\frac{9}{2}$), confirming $9$; treating it as one triangle gives the wrong $\frac{9}{2}$.
_Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.6_
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