Question

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

Answer 1

`A + B + E = 32`

Explanation:

Given

`\int\limits {x^2\cdot e^(-4x)} \, dx = -(1)/(64)e^(-4x)[Ax^2 + Bx + E]C`

Required

Find
`A +B + E`

We have:

`\int\limits {x^2\cdot e^(-4x)} \, dx = -(1)/(64)e^(-4x)[Ax^2 + Bx + E]C`

Using integration by parts

`\int {u} \, dv = uv - \int vdu`

Where

`u = x^2` and
`dv = e^(-4x)dx`

Solve for du (differentiate u)

`du = 2x\ dx`

Solve for v (integrate dv)

`v = -(1)/(4)e^(-4x)`

So, we have:

`\int {u} \, dv = uv - \int vdu`

`\int\limits {x^2\cdot e^(-4x)} \, dx = x^2 *-(1)/(4)e^(-4x) - \int -(1)/(4)e^(-4x) 2xdx`

`\int\limits {x^2\cdot e^(-4x)} \, dx = -(x^2)/(4)e^(-4x) - \int -(1)/(2)e^(-4x) xdx`

`\int\limits {x^2\cdot e^(-4x)} \, dx = -(x^2)/(4)e^(-4x) +(1)/(2) \int xe^(-4x) dx`

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Solving

`\int xe^(-4x) dx`

Integration by parts

`u = x` ----
`du = dx`

`dv = e^(-4x)dx` ----------
`v = -(1)/(4)e^(-4x)`

So:

`\int xe^(-4x) dx = -(x)/(4)e^(-4x) - \int -(1)/(4)e^(-4x)\ dx`

`\int xe^(-4x) dx = -(x)/(4)e^(-4x) + \int e^(-4x)\ dx`

`\int xe^(-4x) dx = -(x)/(4)e^(-4x) -(1)/(4)e^(-4x)`

So, we have:

`\int\limits {x^2\cdot e^(-4x)} \, dx = -(x^2)/(4)e^(-4x) +(1)/(2) \int xe^(-4x) dx`

`\int\limits {x^2\cdot e^(-4x)} \, dx = -(x^2)/(4)e^(-4x) +(1)/(2) [ -(x)/(4)e^(-4x) -(1)/(4)e^(-4x)]`

Open bracket

`\int\limits {x^2\cdot e^(-4x)} \, dx = -(x^2)/(4)e^(-4x) -(x)/(8)e^(-4x) -(1)/(8)e^(-4x)`

Factor out
`e^(-4x)`

`\int\limits {x^2\cdot e^(-4x)} \, dx = [-(x^2)/(4) -(x)/(8) -(1)/(8)]e^(-4x)`

Rewrite as:

`\int\limits {x^2\cdot e^(-4x)} \, dx = [-(1)/(4)x^2 -(1)/(8)x -(1)/(8)]e^(-4x)`

Recall that:

`\int\limits {x^2\cdot e^(-4x)} \, dx = -(1)/(64)e^(-4x)[Ax^2 + Bx + E]C`

`\int\limits {x^2\cdot e^(-4x)} \, dx = [-(1)/(64)Ax^2 -(1)/(64) Bx -(1)/(64) E]Ce^(-4x)`

By comparison:

`-(1)/(4)x^2 = -(1)/(64)Ax^2`

`-(1)/(8)x = -(1)/(64)Bx`

`-(1)/(8) = -(1)/(64)E`

Solve A, B and C

`-(1)/(4)x^2 = -(1)/(64)Ax^2`

Divide by
`-x^2`

`(1)/(4) = (1)/(64)A`

Multiply by 64

`64 * (1)/(4) = A`

`A =16`

`-(1)/(8)x = -(1)/(64)Bx`

Divide by
`-x`

`(1)/(8) = (1)/(64)B`

Multiply by 64

`64 * (1)/(8) = (1)/(64)B*64`

`B = 8`

`-(1)/(8) = -(1)/(64)E`

Multiply by -64

`-64 * -(1)/(8) = -(1)/(64)E * -64`

`E = 8`

So:

`A + B + E = 16 +8+8`

`A + B + E = 32`