Question

If for $n=4$ the approximate value of integral ${\int }_1^9{x}^{2}dx$ by trapezoidal rule is $2\left[\frac{1}{2}\right(1+9^2)+{\alpha }^2+{\beta }^2+7^2]$, then

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

Answer 1

Answer: (C)

$\alpha =3,\beta =5$

$n=4$(given)

${\int }_1^9{x}^{2}dx$ (using trapezoid rule)

$(1-9)$ is divided in $4$ parts

So width, $△x=\frac{b-a}{n}=\frac{9-1}{4}=2$

Using formula,

${\int }_a^{b}f\left(x\right)dx=\frac{△x}{2}\left[f\right(x_0)+2f(x_1)+2f(x_2)...+2f(x_{x-1})+f(x_x\left)\right]$

We get,

${\int }_1^g{x}^{2}dx=2[{\frac{1}{2}}^2+{\left(3\right)}^2+{\left(5\right)}^2+{\left(7\right)}^2+{\frac{9}{2}}^2]$

Comparing with given equation,

$2\left[\frac{1}{2}(1+9^2\right)+{\alpha }^2+{\beta }^2+7^2]$

We get,

$\alpha =3\beta =5$ OR $\alpha =5\beta =3$