Q. 9 - 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.

        What is the actual error that results from a right-sum approximation with $n=4\text{ for }{\int }_1^4 \frac{1}{x}dx?$

Step-by-Step Solution

Answer

Ans:  The actual error is $0.2385$

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. Solution

 $\begin{array}{r}f\left(x\right)=\frac{1}{x} \\ \mathrm{\Delta }x=\frac{b-a}{n}=\frac{4-1}{4}=\frac{3}{4} \\ x_k=a+k\mathrm{\Delta }x=\left(1+\frac{3k}{4}\right) \\ f\left(x_k\right)=f\left(1+\frac{3k}{4}\right) \\ {\int }_1^4 f\left(x\right)dx=\sum _{k=1}^4 f\left(1+\frac{3k}{4}\right)\frac{3}{4} \\ {\int }_1^4 f\left(x\right)dx=\sum _{k=1}^4 \left(\frac{4}{3k+4}\right)\frac{3}{4 } \\ {\int }_1^4 f\left(x\right)dx=\left[\frac{4}{7}+\frac{4}{10}+\frac{4}{13}+\frac{4}{16}\right]\frac{3}{4} \\ {\int }_1^4 f\left(x\right)dx=[0.57+0.4+0.31+0.25]\frac{3}{4} \\ {\int }_1^4 f\left(x\right)dx=\left(1.53\right)\frac{3}{4} \\ {\int }_1^4 f\left(x\right)dx=1.1475 \end{array}$

Since the actual value is ${\int }_1^4 \frac{1}{x}dx=1.386$

Error $=1.386-1.1475$

Thus, the error is $0.2385$.