Question

Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.) f(x) = 8x3/5 + 3x−4/5

Answer 1

Answer:

$\dfrac{8x^{4}}{20}+\dfrac{3x^{2}}{2}-\dfrac{4}{5}x+C$

Step-by-step explanation:

Given the function: $f(x)=\dfrac{8x^3}{5}+3x-\dfrac{4}{5}$

To take the antiderivative (or integral) of a function, we follow the format below.

$f(x)=x^n\\ Then its antiderivative\\Antiderivative of f(x) =\dfrac{x^{n+1}}{n+1}$

Therefore, the antiderivative of f(x) is:

$=\dfrac{8x^{3+1}}{5(3+1)}+\dfrac{3x^{1+1}}{2}-\dfrac{4}{5}x+C\\=\dfrac{8x^{4}}{20}+\dfrac{3x^{2}}{2}-\dfrac{4}{5}x+C$

We want to check our result by differentiation.

$\dfrac{d}{dx}\left(\dfrac{8x^{4}}{20}+\dfrac{3x^{2}}{2}-\dfrac{4}{5}x+C\right)\\=\dfrac{d}{dx}\left(\dfrac{8x^{4}}{20}\right)+\dfrac{d}{dx}\left(\dfrac{3x^{2}}{2}\right)-\dfrac{d}{dx}\left(\dfrac{4}{5}x\right)+\dfrac{d}{dx}\left(C\right)\\\\=\dfrac{32x^{3}}{20}+\dfrac{6x}{2}-\dfrac{4}{5}+0\\\\=\dfrac{8x^{3}}{5}+3x-\dfrac{4}{5}$