Q. 73 - Calculus

Question

Find a function that has the given derivative and value. In each case you can find the answer with an educated guess and check process it may be helpful to do some preliminary algebra

$f\text{'}\left(x\right)=3x^5-2x^2+4 ; f\left(0\right)=1$

Step-by-Step Solution

Answer

The antiderivative can be given as $f\left(x\right)=\frac{x^6}{2}-\frac{2x^3}{3}+4x+1$

Step 1: Given information

We are given the derivative as $f\text{'}\left(x\right)=3x^5-2x^2+4 ; f\left(0\right)=1$

Step 2: Find the antiderivative

We know that differentiating a power function decreases the power by one we can start with the function $f\left(x\right)=x^6-x^3+4x+c$

On differentiating the above function we get,

$f\text{'}\left(x\right)=6x^5-3x^2+4$ which is nearly equal

Now we only adjust the coefficient

We get, $f\left(x\right)=\frac{x^6}{2}-\frac{2x^3}{3}+4x+c$

We are also given that $$\begin{gathered} f\left(0\right)=1 \\ Substittuing this in the function we get, \\ f\left(0\right)=c \\ c=1 \end{gathered}$$

Hence the antiderivative becomes

$f\left(x\right)=\frac{x^6}{2}-\frac{2x^3}{3}+4x+1$

Step 3: Conclusion

The antiderivative can be given as $f\left(x\right)=\frac{x^6}{2}-\frac{2x^3}{3}+4x+1$