Question

The approximate value of ${\int }_1^9{x}^{2}dx$ by using the Trapezoidal rule with $4$ equal intervals is?

• A. $250$
• B. $245$
• C. $242.8$
• D. $243$

Answer 1

The correct option is A $250$

${\int }_1^9{x}^{2}dxb=aa=1\Rightarrow f\left(x\right)=x^2\Rightarrow n=f\left(4\right)\Rightarrow \Delta x=\frac{b-a}{n}=\frac{9-1}{4}=2x_0=1x_1=3x_2=5x_3=7x_4=9{(x}_0{)}^2=1{(x}_1{)}^2=9{(x}_2{)}^2=25{(x}_3{)}^2=49{(x}_4{)}^2=81\Rightarrow {\int }_a^{b}f(x)dx=\frac{\Delta x}{2}\left[f\right(x_0)+2f(x_1).......f(x_4\left)\right]\Rightarrow {\int }_{-1}^9{x}^{2}dx=\frac{2}{2}[1+18+50+98+81]\Rightarrow {\int }_1^9{x}^{2}dx=90+50+110\Rightarrow 250$