Q. 49 - Calculus

Question

Find a function $f$ that has the given derivative $f\text{'}$ and value $f\left(c\right)$. Find an antiderivative of $f\text{'}$ by hand, if possible; if it is not possible to antidifferentiation by hand, use the Second Fundamental Theorem of Calculus to write down an antiderivative.

     $f^{\mathrm{\prime }}\left(x\right)=\frac{1}{2x-1},f\left(1\right)=3$

Step-by-Step Solution

Answer

Ans:    The function is, $f\left(x\right)=\frac{1}{2}(\ln |2x-1\left|\right)+3$

Step 1. Given information.

given,

        $f^{\mathrm{\prime }}\left(x\right)=\frac{1}{2x-1},f\left(1\right)=3$

Step 2. The objective is to find a function f ' meeting the above values.

 Let,

      $\begin{array}{r}y=2x-1\\ dy=2dx\\ dx=\frac{dy}{2}\end{array}$

So,

    $\begin{array}{r}f\left(x\right)=\int f^{\mathrm{\prime }}\left(x\right)dx\\ =\int \frac{1}{y}\frac{dy}{2}\\ =\frac{1}{2}(\ln |y\left|\right)+c\\ =\frac{1}{2}(\ln |2x-1\left|\right)+c\end{array}$

The function, $\frac{1}{2}(\ln |2x-1\left|\right)+c$.

Step 3. Finding the value of c ,

  $\begin{array}{r}f\left(1\right)=3\\ \frac{1}{2}(\ln |2\left(1\right)-1\left|\right)+c=3\\ \frac{1}{2}\ln 1+c=3\\ c=3\end{array}$

Therefore, the function is $f\left(x\right)=\frac{1}{2}(\ln |2x-1\left|\right)+3$