If \[ \int e^x (x^3 + x^2 - x + 4) \, dx = e^x f(x) + C, \] then \( f(1) \) is:
\( 3 \)
Step 1: Understanding the Given Integral We are given: \[ \int e^x (x^3 + x^2 - x + 4) \, dx = e^x f(x) + C \] Using the standard integral formula: \[ \int e^x g(x) \, dx = e^x G(x) + C, \] where \( G(x) \) is the integral of \( g(x) \), we identify: \[ f(x) = \int (x^3 + x^2 - x + 4) \, dx \]
Step 2: Compute \( f(x) \) \[ f(x) = \int (x^3 + x^2 - x + 4) \, dx \] Integrating each term: \[ \int x^3 \, dx = \frac{x^4}{4}, \quad \int x^2 \, dx = \frac{x^3}{3}, \quad \int (-x) \, dx = -\frac{x^2}{2}, \quad \int 4 \, dx = 4x \] Thus, \[ f(x) = \frac{x^4}{4} + \frac{x^3}{3} - \frac{x^2}{2} + 4x \]
Step 3: Evaluate \( f(1) \) \[ f(1) = \frac{(1)^4}{4} + \frac{(1)^3}{3} - \frac{(1)^2}{2} + 4(1) \] \[ = \frac{1}{4} + \frac{1}{3} - \frac{1}{2} + 4 \] Taking LCM (12): \[ = \frac{3}{12} + \frac{4}{12} - \frac{6}{12} + 4 \] \[ = \frac{1}{12} + 4 = 3 \] Thus, \( f(1) = 3 \).
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]

