Q. 8 - 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 antidifferentiate.

           Use the given antiderivative fact to find the exact value of ${\int }_1^4 \frac{1}{x}dx$. (Hint: What is an antiderivative of $\frac{1}{x}$ ? In other words, what is a function F whose derivative is $f\left(x\right)=\frac{1}{x}$ ?)

Step-by-Step Solution

Answer

Ans:  

  $\begin{array}{r}{\int }_1^4 \frac{1}{x}dx=1.386-0\\ {\int }_1^4 \frac{1}{x}dx=1.386 \end{array}$

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

 The antiderivative function $F\left(x\right)$ of the integral ${\int }_1^4 \frac{1}{x}dx\text{ is }F\left(x\right)=\ln \left|x\right|$

So,

    $\begin{array}{r}F\left(1\right)=\ln \left|1\right|=0 \\ F\left(4\right)=\ln \left|4\right|=1.386\end{array}$

Thus,

   $\begin{array}{r}{\int }_1^4 \frac{1}{x}dx=1.386-0\\ {\int }_1^4 \frac{1}{x}dx=1.386 \end{array}$