If p(x) is a cubic polynomial with p(1) = 3, p(0) = 2 and p (–1) = 4, then ∫1−1p(x)dx is
A
2
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B
3
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C
4
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D
5
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Solution
The correct option is D 5 p(x)=ax3+bx2+cx+d p(0) = 2 d = 2 p(1) = 3 a + b + c + 2 = 3 a + b + c = 1 …………… (1) p(-1) = 4 -a +b - c +2 =4 -a +b -c =2 from(1)and (2) 2b=3⇒b=32P(x)=ax3+32x2+cx+2∫1−1p(x)dx=∫1−1(ax2+32x2+cx+2)dx=5