Question

If for $n = 4$, the approximate value of integral $\int_{1}^{9}{x^{2}dx}$ by Trapezoidal rule is $2\left\lbrack \frac{1}{2}(1 + 9^{2}) + \alpha^{2} + \beta^{2} + 7^{2} \right\rbrack$, then

• A. $\alpha = 1,\mspace{6mu}\mspace{6mu}\beta = 3$
• B. $\alpha = 2,\mspace{6mu}\mspace{6mu}\beta = 4$
• C. $\alpha = 3,\mspace{6mu}\mspace{6mu}\beta = 5$
• D. $\alpha = 4,\mspace{6mu}\mspace{6mu}\beta = 6$

Answer 1

Answer: (C)

$\alpha = 3,\mspace{6mu}\mspace{6mu}\beta = 5$

Answer 2

$h = \frac{b - a}{n} = \frac{9 - 1}{4} = 2$

$x_{0} = 1$,$x_{1} = x_{0} + nh = 1 + 1.2 = 3$, $y_{0} = 1,\mspace{6mu}\mspace{6mu} y_{1} = 9,\mspace{6mu}\mspace{6mu} y_{2} = 25,\mspace{6mu}\mspace{6mu} y_{3} = 49,\mspace{6mu}\mspace{6mu} y_{4} = 81$

By trapezoidal rule,

$\int_{a}^{b}{f(x)dx} = \frac{h}{2}\left\lbrack (y_{0} + y_{4}) + 2(y_{1} + y_{2} + y_{3}) \right\rbrack$

$= \frac{2}{2}\left\lbrack (1 + 81) + 2(9 + 25 + 49) \right\rbrack = 2\left\lbrack \frac{1}{2}(1 + 9^{2}) + (3^{2} + 5^{2} + 7^{2}) \right\rbrack$

Obvious from above equation, $\alpha = 3,\mspace{6mu}\mspace{6mu}\beta = 5$.