Q. 51 - 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.

       $f^{\mathrm{\prime }}\left(x\right)=\frac{1}{x^3+1},f\left(2\right)=0$

Step-by-Step Solution

Answer

Ans:  The function is, $f\left(x\right)={\int }_2^x \frac{1}{t^3+1}dt$

Step 1. Given information.

given,

     $f^{\mathrm{\prime }}\left(x\right)=\frac{1}{x^3+1},f\left(2\right)=0$

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

Now, if $f\text{'}$ is continuous on $[a,b]$ then for all $x\in [a,b]$

     $\frac{d}{dx}{\int }_a^x f\left(t\right)dt=f\left(x\right)$

Step 3. Find function,

If $F$ is an antiderivative of $f$ and $f$ is continuous on $[a,b]$ then $F\left(x\right)={\int }_a^x f\left(t\right)dt$ for all $x\in [a,b]$

Using the fact that $f\left(2\right)=0$ the derivative is,

     $f\left(x\right)={\int }_2^x \frac{1}{t^3+1}dt$

Therefore, the function is  $f\left(x\right)={\int }_2^x \frac{1}{t^3+1}dt$