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Q.

∫−23|1−x2|dx=

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An Intiative by Sri Chaitanya

a

283

b

143

c

73

d

13

answer is A.

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Detailed Solution

∫−23|1−x2|dx=∫−2−1(x2−1)dx+∫−11(1−x2)dx+∫13(x2−1)dx =x33|−2−1−1+2−x33|−11+x33|13−2 =73−53+263=283

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∫−23|1−x2|dx=