Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 12 Marks
Question
Evaluate $\int_{-1}^2(|x+1|+|x|+|x-1|) d x$
✓
Answer
Let
$I =\int_{-1}^2(|x+1|+|x|+|x-1|) d x$,
then $ I=\int_{-1}^2|x+1| d x+\int_{-1}^2|x| d x+\int_{-1}^2|x-1| d x$
$=\int_{-1}^2(x+1) d x-\int_{-1}^0 x d x+\int_0^2 x d x-\int_{-1}^1(x-1) d x+\int_1^2(x-1) d x$
$=\left\{\frac{x^2}{2}+x\right\}_{-1}^2-\left\{\frac{x^2}{2}\right\}_{-1}^0+\left\{\frac{x^2}{2}\right\}_0^2-\left\{\frac{x^2}{2}-x\right\}_{-1}^1+\left\{\frac{x^2}{2}-x\right\}_1^2$
$=\left\{(4)-\left(-\frac{1}{2}\right)\right\}-\left\{-\frac{1}{2}\right\}+\{2\}-\left\{\left(-\frac{1}{2}\right)-\left(\frac{3}{2}\right)\right\}+\left\{(0)-\left(-\frac{1}{2}\right)\right\}$
$=\left\{4+\frac{1}{2}\right\}+\left\{\frac{1}{2}\right\}+\{2\}+\{2\}+\left\{\frac{1}{2}\right\}=\frac{19}{2}$
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