Chapter 6, Problem 8RE

To determine

To find :the value of sumswhich are converging.

Answer to Problem 8RE

The value is: $1.60944$

Explanation of Solution

Given information:

To use the fundamental theorem of calculus.

Calculation:

If $f$ is continuous at every point of $[a,b],$ and if $F$ is any anti-derivative of $f$ on $[a,b],$ then

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

This part of the fundamental theorem is also called the integral evaluation theorem.

Proof:

Part 1 of the fundamental theorem tells us that an anti-derivative of $f$ exists, namely

  $G(x)={{\displaystyle \int }}_a^{x}f(t)dt$

Thus, if $F$ is any anti-derivative of $f$ then for some constant $F(x)=G(x)+C$ for some constant $C$

  $\begin{array}{l}\begin{array}{cc}F(b)-F(a)& =[G(b)+C]-[G(a)+C]\\ & =G(b)-G(a)\\ & ={{\displaystyle \int }}_a^{b}f(t)dt-{{\displaystyle \int }}_a^{a}f(t)dt\\ & ={{\displaystyle \int }}_a^{b}f(t)dt-0\\ & \end{array}\\ F(b)-F(a)={{\displaystyle \int }}_a^{b}f(t)dt\end{array}$

So,

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

Now, use the above proof to find the value of integral.

  $f(x)=\frac{1}{x}$

If $F$ is any anti-derivative of $f$ then

  $F(x)=\ln x$

So

  $\begin{array}{cc}{{\displaystyle \int }}_1^5\left(\frac{1}{x}\right)dx& =[\ln 5]-[\ln 1]\\ & \approx 1.60944-0\end{array}$

= $1.60944$

Hence, the value is $1.60944$