Question

Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative. Remember to use absolute values where appropriate.) f(x) = x5 - x3 + 8x - x4

Answer 1

Answer:

$\frac{x^6}{6}- \frac{x^5}{5}- \frac{x^4}{4} +4x^2+C$

Step-by-step explanation:

Step 1

The first step is to rearrange the function so that powers of x are in descending form as show below,

$f(x)=x^5-x^3+8x-x^4=x^5-x^4-x^3+8x$

Step 2

The second step is to realize that the anti derivative is the indefinite integral of the function . We compute the integral and then add the constant C in the end. We will use the result that

$\int x^ndx=\frac{x^{n+1}}{n+1} +C$ on every term of the fraction.

This calculation is shown below,

$\int f(x)dx=\int \bigl( x^5-x^4-x^3+8x\bigr)dx\\\int f(x)dx=\bigl(\frac{x^{5+1}}{5+1} -\frac{x^{4+1}}{4+1}- \frac{x^{3+1}}{3+1}+\frac{8x^{1+1}}{(1+1)}+C \bigr)\\\int f(x)dx= \frac{x^6}{6}- \frac{x^5}{5}- \frac{x^4}{4} +4x^2+C$

Step 3

The third step is to verify using differentiation that the indefinite integral is correct. We will do this by differentiating every term of the integral with respect to x as shown below

$\frac{d}{dx} \bigl[\frac{x^6}{6} -\frac{x^5}{5}- \frac{x^4}{4} +4x^2+c\bigr]=6\cdot\frac{x^{6-1}}{6}-5\cdot\frac{x^{5-1}}{5}-4\cdot\frac{x^{4-1}}{4}+2\cdot 4x^{2-1}+0=x^5-x^4-x^3+8x$

This shows that the integral is indeed correct.