Question

lf ${\log }_{10}100=2,{\log }_{10}101=2.004$, ${\log }_{10}102=2.0086,{\log }_{10}103=2.0128$ then ${\int }_{100}^{103}{\log }_{10}xdx$ by Trapezoidal rule is?

• A. $6.019$
• B. $6.0019$
• C. $6.1093$
• D. $6.11993$

Answer 1

The correct option is A $6.019$

${\int }_a^{b}f\left(x\right)dx=\frac{b-a}{2N}\left[f\right(x_1)+2f(x_2)+2f(x_3)+....f(x_{N+1}\left)\right]$

$\Rightarrow {\int }_{100}^{103}{\log }_{10}xdx=\frac{3}{2\times 3}[2+2(2.004)+2(2.0086)+2.0128]$

$=[1+2.004+2.0086+1.0064]$

$=6.019$