Question

If $\int e^x (x^3 + x^2 - x + 4) dx = e^x f(x) + c$, then $f(1) =$

Answer 1

3

Answer 2

The given integral is $\int e^x (x^3 + x^2 - x + 4) dx$. We are given that the integral is equal to $e^x f(x) + c$. This integral is of the form $\int e^x P(x) dx$, where $P(x)$ is a polynomial. A standard result for such integrals is $\int e^x P(x) dx = e^x (P(x) - P'(x) + P''(x) - P'''(x) + \dots) + c$, where the derivatives are taken until they become zero.

In this case, $P(x) = x^3 + x^2 - x + 4$. Let's find the derivatives of $P(x)$:

$P'(x) = \frac{d}{dx}(x^3 + x^2 - x + 4) = 3x^2 + 2x - 1$

$P''(x) = \frac{d}{dx}(3x^2 + 2x - 1) = 6x + 2$

$P'''(x) = \frac{d}{dx}(6x + 2) = 6$

$P^{(4)}(x) = \frac{d}{dx}(6) = 0$

Using the formula, the integral is:

$\int e^x (x^3 + x^2 - x + 4) dx = e^x (P(x) - P'(x) + P''(x) - P'''(x)) + c$

$= e^x ((x^3 + x^2 - x + 4) - (3x^2 + 2x - 1) + (6x + 2) - (6)) + c$

$= e^x (x^3 + x^2 - x + 4 - 3x^2 - 2x + 1 + 6x + 2 - 6) + c$

Combine the terms inside the parenthesis:

$= e^x (x^3 + (1 - 3)x^2 + (-1 - 2 + 6)x + (4 + 1 + 2 - 6)) + c$

$= e^x (x^3 - 2x^2 + 3x + 1) + c$

We are given that $\int e^x (x^3 + x^2 - x + 4) dx = e^x f(x) + c$. Comparing this with our result, we find that $f(x) = x^3 - 2x^2 + 3x + 1$.

We need to find the value of $f(1)$. Substitute $x = 1$ into the expression for $f(x)$:

$f(1) = (1)^3 - 2(1)^2 + 3(1) + 1$

$f(1) = 1 - 2(1) + 3(1) + 1$

$f(1) = 1 - 2 + 3 + 1$

$f(1) = 5 - 2$

$f(1) = 3$