If the integral of the product of x squared and e raised to the negative 4 times x power, dx equals the product of negative 1 over 64 times e raised to the negative 4 times x power and the quantity A times x squared plus B times x plus E, plus C , then the value of A B E is

Question

01-06-2021
Mathematics

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If the integral of the product of x squared and e raised to the negative 4 times x power, dx equals the product of negative 1 over 64 times e raised to the negative 4 times x power and the quantity A times x squared plus B times x plus E, plus C , then the value of A B E is

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MrRoyal MrRoyal · Answer 1 · 2021-06-03T13:10:11+02:00

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]