Which of the following values of a, b, c and d will produce a quadrature formula

\(\displaystyle\int_{−1}^1\) f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)

that has degree of precision 3?

Explanation:

Given 

\(\displaystyle\int_{−1}^1\) f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)....(i)

Let f(x) = 1 then from (i)

\(\displaystyle\int_{−1}^1\)dx = a + b 

⇒ 2 = a + b...(ii)

For f(x) = x ⇒ f'(x) = 1 so from (i)

\(\displaystyle\int_{−1}^1\)x dx = -a + b + c + d

0 = -a + b + c + d....(iii)

For f(x) = x² ⇒ f'(x) = 2x so from (i)

\(\displaystyle\int_{−1}^1\)x² dx = a + b - 2c + 2d

\(\frac23\) = a + b - 2c + 2d....(iv)

and for f(x) = x³ ⇒ f'(x) = 3x² so from (i)

\(\displaystyle\int_{−1}^1\)x3 dx = -a + b + 3c + 3d

0 = -a + b + 3c + 3d....(v)

Multiply (iii) by 3 we get

-3a  +3b + 3c + 3d = 0...(vi)

Subtract (v) and (vi) we get

2a - 2b = 0 ⇒ a = b...(vii)

Putting a = b in (ii)

2a = 2 ⇒ a = 1

Hence a = b = 1

Putting these values of a and b in (iii) and (iv) we get

c + d = 0 ...(viii) and 

2 - 2c + 2d = \(\frac23\)

2c - 2d = \(\frac43\) ⇒ c - d = \(\frac23\) ...(ix)

adding (viii) and (ix) we get

2c = \(\frac23\) so c = \(\frac13\)

Hence d = - \(\frac13\)    

Therefore we get a = 1, b = 1, c = \(\frac{1}{3}\), d = \(−\frac{1}{3}\)

Option (1) is correct