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 role="math" localid="1649787152886" ${\int }_a^b f\left(x\right)dx=F\left(b\right)-F\left(a\right)$, where $F$is any antiderivative of role="math" localid="1649787141650" $f$. 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 $n=4\text{for}{\int }_1^4 \frac{1}{x}dx?$

Short answer

Ans: The actual error is $0.2385$

Step-by-step solution

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$0.2385$.