Question

determine the following.\
\\(\int 7e^{-0.4x} \\, dx\\)\
\\(\int 7e^{-0.4x} \\, dx = \square\\) (type an exact answer.)

Explanation:

Step1: Use Constant Multiple Rule

The integral of a constant multiple of a function is the constant multiple of the integral of the function. So, $\int 7e^{-0.4x} dx = 7\int e^{-0.4x} dx$.

Step2: Use Exponential Integral Rule

The integral of $e^{ax} dx$ (where $a$ is a constant) is $\frac{1}{a}e^{ax} + C$. Here, $a = -0.4$, so $\int e^{-0.4x} dx = \frac{1}{-0.4}e^{-0.4x} + C$.

Step3: Multiply by the Constant

Multiply the result from Step 2 by 7: $7\times\frac{1}{-0.4}e^{-0.4x} + C$. Simplify $\frac{7}{-0.4}$: $7\div(-0.4)=-\frac{7}{0.4}=-\frac{70}{4}=-\frac{35}{2}$. So the integral is $-\frac{35}{2}e^{-0.4x} + C$.

Answer:

$-\frac{35}{2}e^{-0.4x} + C$ (or equivalent forms, e.g., $-17.5e^{-0.4x} + C$)