Q. 50 - 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.

        $f^{\mathrm{\prime }}\left(x\right)=\frac{1}{x^2+1},f(-1)=0$

Step-by-Step Solution

Answer

Ans:   The function is, $f\left(x\right)={\tan }^{-1}x+\frac{\pi }{4}$

Step 1. Given information.

given,

        $f^{\mathrm{\prime }}\left(x\right)=\frac{1}{x^2+1},f(-1)=0$

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

So,

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

The function is ${\tan }^{-1}x+c$

Step 3. Finding the value of c ,

   $\begin{array}{r}f(-1)=0\\ {\tan }^{-1}(-1)+c=0\\ -\frac{\pi }{4}+c=0\\ c=\frac{\pi }{4}\end{array}$

Therefore, the function is $f\left(x\right)={\tan }^{-1}x+\frac{\pi }{4}$.