A) ∫−11f(x)dx=1
∫−11k(3−x2)dx=(3kx−kx3/3)∣−11=3k−k/3+3k−k/3=6k−2k/3=18k/3−2k/3=16k/3=1
k=3/16
B)∫01(3(3−x2)/16)dx=(9x/16−x3/16)∣01=9/16−1/16=8/16=1/2
C)∫−1/21/2(3(3−x2)/16)dx=(9x/16−x3/16)∣−1/21/2=9/32−1/128+9/32−1/128=18/32−2/128=9/16−1/64=36/64−1/64=35/64
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