e-x(4x-9)+C
e-x(4x-1)+C
We are given the integral \( \int (5 - 4x)e^{-x} \, dx \) and asked to find the solution.
We can solve this using the method of integration by parts. Let:
\( u = 5 - 4x \), so that \( du = -4 \, dx \),
and let \( dv = e^{-x} \, dx \), so that \( v = -e^{-x} \).
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:
\( \int (5 - 4x) e^{-x} \, dx = -(5 - 4x) e^{-x} - \int (-e^{-x})(-4) \, dx \).
Simplifying this, we get:
\( -(5 - 4x) e^{-x} + 4 \int e^{-x} \, dx \).
The integral of \( e^{-x} \) is \( -e^{-x} \), so we have:
\( -(5 - 4x) e^{-x} - 4e^{-x} + C \),
which simplifies to:
\( e^{-x}(4x - 9) + C \).
The correct answer is \( e^{-x}(4x - 9) + C \).
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to: