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If $p(x)$ is a cubic polynomial with $p(1)=3, p(0)=2$ and $p(-1)=4$, then $\int_{-1}^{1} p(x) d x$ is
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The correct answer is:
5
$\mathrm{p}(\mathrm{x})=\mathrm{ax}^{3}+\mathrm{b} \mathrm{x}^{2}+\mathrm{cx}+\mathrm{d}$
$\mathrm{p}(0)=2$
$\mathrm{~d}=2$
$\mathrm{p}(1)=3$
$\mathrm{a}+\mathrm{b}+\mathrm{c}+2=3$
$\mathrm{a}+\mathrm{b}+\mathrm{c}=1$
$\mathrm{p}(-1)=4$
$-a+b-c+2=4$
$-a+b-c=2$
$2 b=3 \Rightarrow b=\frac{3}{2}$
$P(x)=a x^{3}+\frac{3}{2} x^{2}+c x+2$
$\int_{-1}^{1} p(x) d x=\int_{-1}^{1}\left(a x^{2}+\frac{3}{2} x^{2}+c x+2\right) d x=5$
$\mathrm{p}(0)=2$
$\mathrm{~d}=2$
$\mathrm{p}(1)=3$
$\mathrm{a}+\mathrm{b}+\mathrm{c}+2=3$
$\mathrm{a}+\mathrm{b}+\mathrm{c}=1$
$\mathrm{p}(-1)=4$
$-a+b-c+2=4$
$-a+b-c=2$
$2 b=3 \Rightarrow b=\frac{3}{2}$
$P(x)=a x^{3}+\frac{3}{2} x^{2}+c x+2$
$\int_{-1}^{1} p(x) d x=\int_{-1}^{1}\left(a x^{2}+\frac{3}{2} x^{2}+c x+2\right) d x=5$
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