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Question

If f(x)=a+bx+cx2, show that
10f(x)dx=16[f(0)+4f(12)+f(1)]

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Solution

f(x)=a+bx+cx2
f(0)=a
f(1/2)=a+b2+c4
f(1)=a+b+c
10f(x)dx=10(a+bx+cx2)dx
=ax+bx22+cx33|10
=a(10)+b2(10)+c3(10)
=a+b2+c3......(1)LHS
RHS=16[f(0)+4f(1/2)+f(1)]
=16[a+4(a+b2+c4)+a+b+c]
=16(a+4a+a+2b+c+b+c)
=a+3b6+2c6
=a+b2+c3RHS
LHS=RHS
Hence proved.

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