Question

Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.)
f(x) =

2x4 + 4x3 − x
x3
, x > 0

F(x) =

2) Find f. (Use C for the constant of the first antiderivative and D for the constant of the second antiderivative.)

f ''(x) = 32x3 − 18x2 + 8x

f(x) =

Answer 1

Answer:

1) F(x)=$\frac{2x^{5} }{5}+x^{4}-\frac{x^{2} }{2}+C$

2) $f(x)=\frac{8x^5}{5}-\frac{3x^4}{2}+\frac{4x^3}{3}+D$

Step-by-step explanation:

Applying the antiderivation rules:

F(x)= $\frac{2x^{4+1} }{4+1}+4\frac{x^{3+1} }{3+1}-\frac{x^{1+1} }{1+1}+C$

F(x)=$\frac{2x^{5} }{5}+x^{4}-\frac{x^{2} }{2}+C$

Checking by differentitation we have:

$F'(x)=2*\frac{5x^{4} }{5}+4*\frac{4x^3}{4}+2*\frac{x}{2}\\F'(x)=2x^4+4x^3-x=f(x)$

Which is demonstrated.

2) To find f we must antiderivate twice:

$f'(x)=\int\limits {(32x^3-18x^2+8x)} \, dx\\f'(x)=8x^4-6x^3+4x^2+C\\f(x)=\int\limits( {8x^4-6x^3+4x^2)} \, dx\\ f(x)=8\frac{x^5}{5}-6\frac{x^5}{5}+4\frac{x^3}{3}+D\\ f(x)=\frac{8x^5}{5}-\frac{3x^4}{2}+\frac{4x^3}{3}+D$