Q. 11 - Calculus

Question

We will see that definite integral can be computed by taking differences of antiderivatives; in particular, the Fundamental Theorem of Calculus will reveal that if $f$ is continuous on $[a,b]$, then ${\int }_a^b f\left(x\right)dx=F\left(b\right)-F\left(a\right)$, where F is any antiderivative of $f$. Armed with this fact, we can check the exact error of Riemann sum approximations for integrals of functions that we can antidifferentiation.

  Repeat the preceding steps, but with the midpoint sum in place of the right sum.

 

Step-by-Step Solution

Answer

Ans:  ${\int }_1^4 f\left(x\right)dx=1.297.$ 

Step 1. Given information.

given, 

  ${\int }_a^b f\left(x\right)dx=F\left(b\right)-F\left(a\right)$

Step 2. Since the sum of areas of the right rectangles gives,

 $\begin{array}{r}f\left(x\right)=\frac{1}{x} \\ \mathrm{\Delta }x=\frac{b-a}{n}=\frac{4-1}{12}=\frac{3}{12}=\frac{1}{4} \\ x_k=a+k\mathrm{\Delta }x=\left(1+\frac{k}{4}\right) \\ f\left(x_i\right)=f\left(1+\frac{k}{4}\right) \\ {\int }_1^4 f\left(x\right)dx=\sum _{k=1}^{12} f\left(1+\frac{k}{4}\right)\frac{1}{4} \\ {\int }_1^4 f\left(x\right)dx=\sum _{i=1}^{12} \left(\frac{4}{k+4}\right)\frac{1}{4} \\ {\int }_1^4 f\left(x\right)dx=\left[\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\right] \\ {\int }_1^4 f\left(x\right)dx=\left[\frac{935059}{720720}\right] \\ {\int }_1^4 f\left(x\right)dx=1.297. \end{array}$