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Question: If I = \(\int_{- 3}^{2}{(|x + 1| + ⥂ |x + 2|} + |x - 1|)dx\), then I equals:...

If I = 32(x+1+x+2+x1)dx\int_{- 3}^{2}{(|x + 1| + ⥂ |x + 2|} + |x - 1|)dx, then I equals:

A

312\frac{31}{2}

B

352\frac{35}{2}

C

372\frac{37}{2}

D

392\frac{39}{2}

Answer

312\frac{31}{2}

Explanation

Solution

We can write

I = I1 + I2 + I3

where I1 = 32x+1\int_{- 3}^{2}{|x + 1|}dx etc.

Put x + 1 = t, so that

I1 = 23tdt\int_{- 2}^{3}{|t|}dt =20(t)dt\int_{- 2}^{0}{( - t)}dt + 03tdt\int_{0}^{3}tdt

= –  12t2]20\left. \ \frac{1}{2}t^{2} \right\rbrack_{- 2}^{0} +  12t2]03\left. \ \frac{1}{2}t^{2} \right\rbrack_{0}^{3}= 132\frac{13}{2}

Similarly, I2 = I3 = 92\frac{9}{2}

Thus, I = 312\frac{31}{2}