Question Number 25472 by rita1608 last updated on 10/Dec/17
$${slolve}\:{the}\:{definite}\:{integral}\int_{\mathrm{1}^{} } ^{\mathrm{2}} \frac{\mathrm{1}}{{x}}{dxusingg}>\:\:\:{ng}\: \\ $$ $${trapezoidal}\:{rule}\:{with}\:\mathrm{4}\:{sub}\:{intervals}\: \\ $$ $${hencefind}\:{an}\:{approximate}\:{value}\:{of}\: \\ $$ $${ln}\:\mathrm{2}. \\ $$
Answered by prakash jain last updated on 10/Dec/17
![Four subintervals Δx=((2−1)/4)=(1/4) x_0 =1,x_1 =1.25,x_2 =1.5, x_3 =1.75,x_4 =2 Δx=0.25 I=((Δx)/2)[f(x_0 )+2f(x_1 )+2f(x_2 )+2f(x_3 )+f(x_4 )] =((0.25)/2)[1+(2/(1.25))+(2/(1.5))+(2/(1.75))+(1/2)] =(1/(2×4))[1+((2×4)/5)+((2×2)/3)+((2×4)/7)+(1/2)] =(1/8)[1+1.6+1.33+1.14+.5] =((5.57)/8)=0.69 ∫_1 ^2 (1/x)dx=[ln x]_1 ^2 =ln 2 ln 2≈0.69](Q25473.png)
$$\mathrm{Four}\:\mathrm{subintervals} \\ $$ $$\Delta{x}=\frac{\mathrm{2}−\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ $${x}_{\mathrm{0}} =\mathrm{1},{x}_{\mathrm{1}} =\mathrm{1}.\mathrm{25},{x}_{\mathrm{2}} =\mathrm{1}.\mathrm{5},\:{x}_{\mathrm{3}} =\mathrm{1}.\mathrm{75},{x}_{\mathrm{4}} =\mathrm{2} \\ $$ $$\Delta{x}=\mathrm{0}.\mathrm{25} \\ $$ $${I}=\frac{\Delta{x}}{\mathrm{2}}\left[{f}\left({x}_{\mathrm{0}} \right)+\mathrm{2}{f}\left({x}_{\mathrm{1}} \right)+\mathrm{2}{f}\left({x}_{\mathrm{2}} \right)+\mathrm{2}{f}\left({x}_{\mathrm{3}} \right)+{f}\left({x}_{\mathrm{4}} \right)\right] \\ $$ $$=\frac{\mathrm{0}.\mathrm{25}}{\mathrm{2}}\left[\mathrm{1}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{25}}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{75}}+\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}×\mathrm{4}}\left[\mathrm{1}+\frac{\mathrm{2}×\mathrm{4}}{\mathrm{5}}+\frac{\mathrm{2}×\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}×\mathrm{4}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$ $$=\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{1}+\mathrm{1}.\mathrm{6}+\mathrm{1}.\mathrm{33}+\mathrm{1}.\mathrm{14}+.\mathrm{5}\right] \\ $$ $$=\frac{\mathrm{5}.\mathrm{57}}{\mathrm{8}}=\mathrm{0}.\mathrm{69} \\ $$ $$\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{1}}{{x}}{dx}=\left[\mathrm{ln}\:{x}\right]_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{ln}\:\mathrm{2} \\ $$ $$\mathrm{ln}\:\mathrm{2}\approx\mathrm{0}.\mathrm{69} \\ $$