Let ∫ x³sin x dx=g(x)+C, where C is the constant of integration. If 8(g(π/2)+g^'(π/2))=απ³+βπ²+γ,α,β,γ∈ Z, then α+β-γ equals :

Solution

Maths sir · 4 steps

Get g by the tabular method; g' is just the integrand x³sin x, so it costs you nothing.

1. Tabular method. Keep differentiating x³ until it becomes 0, and keep integrating sin x. Multiply along the diagonals with alternating signs +, -, +, -.

   Sign	Derivatives of x³	Integrals of sin x	Term
   +	x³	-cos x	-x³cos x
   -	3x²	-sin x	+3x²sin x
   +	6x	cos x	+6x cos x
   -	6	sin x	-6 sin x

   g(x) = -x³cos x + 3x²sin x + 6x cos x - 6 sin x

2. Put x = π/2. Here cos π/2 = 0 and sin π/2 = 1, so g(π/2) = 3π²/4 - 6. Since g'(x) = x³sin x, we get g'(π/2) = π³/8.
3. Multiply by 8. 8(π³/8 + 3π²/4 - 6) = π³ + 6π² - 48. So α = 1, β = 6, γ = -48.
4. So. α + β - γ = 1 + 6 + 48 = 55, which is option (b).

The idea to keep. For a polynomial times sin or cos, use the tabular method with alternating signs. And whenever a question asks for g' of an antiderivative, write down the integrand straight away.

Answer from the NTA final answer key. Solution by PrepWiser.