Question

If f(x) = `int(3x - 1)x(x + 1)(18x^11 + 15x^10 - 10x^9)^(1/6)dx`, where f(0) = 0, is in the form of `((18x^α + 15x^β - 10x^γ)^δ)/θ`, then (3α + 4β + 5γ + 6δ + 7θ) is ______. (Where δ is a rational number in its simplest form)

Options

• 296

• 297

• 298

• 299

Answer 1

If f(x) = `int(3x - 1)x(x + 1)(18x^11 + 15x^10 - 10x^9)^(1/6)dx`, where f(0) = 0, is in the form of `((18x^α + 15x^β - 10x^γ)^δ)/θ`, then (3α + 4β + 5γ + 6δ + 7θ) is 298. (Where δ is a rational number in its simplest form)

Explanation:

f(x) = `int(3x - 1)x^2(x + 1)(18x^5 + 15x^4 - 10x^3)^(1/6)dx`

= `int(3x^4 + 2x^3 - x^2)(18x^5 + 15x^4 - 10x^3)^(1/6)dx`

Let 18x⁵ + 15x⁴ – 10x³ = t

(90x⁴ + 60x³ – 30x²)dx = dt

30(3x⁴ + 2x³ – x²)dx = dt

f(x) = `int(dt)/30t^(1/6)`

= `t^(7/6)/30.7. 6 + c`

= `t^(7/6)/35 + c`

f(x) = `(18x^5 + 15x^4 - 10x^3)^(7/6)/35 + c`

α = 5, β = 4, γ = 3, δ = `7/6`, θ = 35

3α + 4β + 5γ + 6δ + 7θ

⇒ 15 + 16 + 15 + 7 + 245 = 298