\( f ^ { \prime } ( x ) = 2 x ^ { 2 } + 5 x - 3 \)

To find the original function \( f(x) \) from its derivative \( f'(x) \), you'll need to perform **antiderivation** (also known as **integration**) on \( f'(x) \). Given: \[ f'(x) = 2x^{2} + 5x - 3 \] ### Step-by-Step Integration 1. **Integrate each term separately:** \[ \int f'(x) \, dx = \int (2x^{2}) \, dx + \int (5x) \, dx + \int (-3) \, dx \] 2. **Apply the power rule of integration:** The power rule states that: \[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \] where \( n \neq -1 \) and \( C \) is the constant of integration. 3. **Integrate each term:** - For \( 2x^{2} \): \[ \int 2x^{2} \, dx = 2 \cdot \frac{x^{3}}{3} = \frac{2}{3}x^{3} \] - For \( 5x \): \[ \int 5x \, dx = 5 \cdot \frac{x^{2}}{2} = \frac{5}{2}x^{2} \] - For \( -3 \): \[ \int (-3) \, dx = -3x \] 4. **Combine the integrals and add the constant of integration:** \[ f(x) = \frac{2}{3}x^{3} + \frac{5}{2}x^{2} - 3x + C \] Here, \( C \) represents the **constant of integration**, accounting for any constant value that was "lost" during differentiation. ### Final Answer \[ f(x) = \frac{2}{3}x^{3} + \frac{5}{2}x^{2} - 3x + C \] Where: - \( \frac{2}{3}x^{3} \) is the antiderivative of \( 2x^{2} \) - \( \frac{5}{2}x^{2} \) is the antiderivative of \( 5x \) - \( -3x \) is the antiderivative of \( -3 \) - \( C \) is the constant of integration If you have any initial conditions or specific values for \( f(x) \) at a particular point, you can solve for the constant \( C \).