Answer :
To find the general antiderivative of given functions, we perform integration. The antiderivative is essentially the reverse of differentiation.
For [tex]f(x) = 5[/tex]:
The antiderivative of a constant [tex]c[/tex] is [tex]cx + C[/tex] where [tex]C[/tex] is the constant of integration.
[tex]\int 5 \, dx = 5x + C[/tex]
For [tex]f(x) = x^2 + \pi[/tex]:
Integrate each term separately. The antiderivative of [tex]x^n[/tex] is [tex]\frac{x^{n+1}}{n+1}[/tex].
[tex]\int (x^2 + \pi) \, dx = \int x^2 \, dx + \int \pi \, dx = \frac{x^3}{3} + \pi x + C[/tex]
For [tex]f(x) = x^{\frac{5}{4}}[/tex]:
Using the power rule:
[tex]\int x^{\frac{5}{4}} \, dx = \frac{x^{\frac{9}{4}}}{\frac{9}{4}} + C = \frac{4}{9}x^{\frac{9}{4}} + C[/tex]
For [tex]f(x) = \frac{1}{\sqrt{x^3}}[/tex]:
Rewrite as [tex]x^{-\frac{3}{2}}[/tex] and integrate:
[tex]\int x^{-\frac{3}{2}} \, dx = \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} + C = -2x^{-\frac{1}{2}} + C = -\frac{2}{\sqrt{x}} + C[/tex]
For [tex]f(x) = x^2 - x[/tex]:
Integrate each term:
[tex]\int (x^2 - x) \, dx = \int x^2 \, dx - \int x \, dx = \frac{x^3}{3} - \frac{x^2}{2} + C[/tex]
For [tex]f(x) = 4x^5 - x^3[/tex]:
Integrate each term:
[tex]\int (4x^5 - x^3) \, dx = 4 \int x^5 \, dx - \int x^3 \, dx = \frac{4x^6}{6} - \frac{x^4}{4} + C = \frac{2}{3}x^6 - \frac{x^4}{4} + C[/tex]
- For [tex]f(x) = 27x^7 + 3x^5 - 45x^3[/tex]:
Integrate each term:
[tex]\int (27x^7 + 3x^5 - 45x^3) \, dx = \frac{27x^8}{8} + \frac{3x^6}{6} - \frac{45x^4}{4} + C = \frac{27}{8}x^8 + \frac{1}{2}x^6 - \frac{45}{4}x^4 + C[/tex]
- For [tex]f(x) = x^2(x^3 + 5x^2 - 3x + 2)[/tex]:
First, expand the expression:
[tex]f(x) = x^5 + 5x^4 - 3x^3 + 2x^2[/tex]
Now integrate each term:
[tex]\int (x^5 + 5x^4 - 3x^3 + 2x^2) \, dx = \frac{x^6}{6} + 5 \cdot \frac{x^5}{5} - \frac{3x^4}{4} + \frac{2x^3}{3} + C = \frac{x^6}{6} + x^5 - \frac{3}{4}x^4 + \frac{2}{3}x^3 + C[/tex]
In each case, the constant [tex]C[/tex] represents the arbitrary constant of integration.
