Respuesta :
Answer:
[tex]A + B + E = 32[/tex]
Step-by-step explanation:
Given
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C[/tex]
Required
Find [tex]A +B + E[/tex]
We have:
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C[/tex]
Using integration by parts
[tex]\int {u} \, dv = uv - \int vdu[/tex]
Where
[tex]u = x^2[/tex] and [tex]dv = e^{-4x}dx[/tex]
Solve for du (differentiate u)
[tex]du = 2x\ dx[/tex]
Solve for v (integrate dv)
[tex]v = -\frac{1}{4}e^{-4x}[/tex]
So, we have:
[tex]\int {u} \, dv = uv - \int vdu[/tex]
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = x^2 *-\frac{1}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x} 2xdx[/tex]
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} - \int -\frac{1}{2}e^{-4x} xdx[/tex]
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx[/tex]
-----------------------------------------------------------------------
Solving
[tex]\int xe^{-4x} dx[/tex]
Integration by parts
[tex]u = x[/tex] ---- [tex]du = dx[/tex]
[tex]dv = e^{-4x}dx[/tex] ---------- [tex]v = -\frac{1}{4}e^{-4x}[/tex]
So:
[tex]\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x}\ dx[/tex]
[tex]\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} + \int e^{-4x}\ dx[/tex]
[tex]\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}[/tex]
So, we have:
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx[/tex]
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} [ -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}][/tex]
Open bracket
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} -\frac{x}{8}e^{-4x} -\frac{1}{8}e^{-4x}[/tex]
Factor out [tex]e^{-4x}[/tex]
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{x^2}{4} -\frac{x}{8} -\frac{1}{8}]e^{-4x}[/tex]
Rewrite as:
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{4}x^2 -\frac{1}{8}x -\frac{1}{8}]e^{-4x}[/tex]
Recall that:
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C[/tex]
[tex]\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{64}Ax^2 -\frac{1}{64} Bx -\frac{1}{64} E]Ce^{-4x}[/tex]
By comparison:
[tex]-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2[/tex]
[tex]-\frac{1}{8}x = -\frac{1}{64}Bx[/tex]
[tex]-\frac{1}{8} = -\frac{1}{64}E[/tex]
Solve A, B and C
[tex]-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2[/tex]
Divide by [tex]-x^2[/tex]
[tex]\frac{1}{4} = \frac{1}{64}A[/tex]
Multiply by 64
[tex]64 * \frac{1}{4} = A[/tex]
[tex]A =16[/tex]
[tex]-\frac{1}{8}x = -\frac{1}{64}Bx[/tex]
Divide by [tex]-x[/tex]
[tex]\frac{1}{8} = \frac{1}{64}B[/tex]
Multiply by 64
[tex]64 * \frac{1}{8} = \frac{1}{64}B*64[/tex]
[tex]B = 8[/tex]
[tex]-\frac{1}{8} = -\frac{1}{64}E[/tex]
Multiply by -64
[tex]-64 * -\frac{1}{8} = -\frac{1}{64}E * -64[/tex]
[tex]E = 8[/tex]
So:
[tex]A + B + E = 16 +8+8[/tex]
[tex]A + B + E = 32[/tex]
