Q. 10 - 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.

  By trial and error with smaller values of n, find the smallest value of n for which a right sum will approximate ${\int }_1^4 \frac{1}{x}dx\text{ to within }0.25$. 

Step-by-Step Solution

Answer

Ans:   $\mathrm{\Delta }x=\frac{3}{n}=0.25n=12$

Step 1. Given information.

given,

       The anti-derivative of the integral is ${\int }_a^b f\left(x\right)dx=F\left(b\right)-F\left(a\right)$

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

 $\begin{array}{r}f\left(x\right)=\frac{1 }{x} \\ \mathrm{\Delta }x=\frac{b-a}{n}=\frac{4-1}{n}=\frac{3}{n} \\ x_k=a+k\mathrm{\Delta }x=\left(1+\frac{3k}{n}\right) \\ f\left(x_k\right)=f\left(1+\frac{3k}{n}\right) \\ {\int }_1^4 f\left(x\right)dx=\sum _{k=1}^n f\left(1+\frac{3k}{n}\right)\frac{3}{n} \end{array}$

Then, 

    $\mathrm{\Delta }x=\frac{3}{n}=0.25n=12$