Let f(θ ) = sin θ + ∫_(- π /2)^(π /2) (sin θ + t cos θ )f(t)dt . Then the value of | ∫₀^(π /2) f(θ )dθ | is _____________.

Solution

Maths sir · 5 steps

Both integrals in t are just constants, so name them A and B and f becomes a sine-cosine combination.

1. Pull out constants. Split the integral as sin θ ∫ f(t) dt + cos θ ∫ t f(t) dt, both over [-π/2, π/2]. Call these numbers A and B. Then f(θ) = (1+A)sin θ + B cos θ.
2. Find A. A = ∫_(-π/2)^(π/2) [(1+A)sin t + B cos t] dt. The sin t part is odd and gives 0, and ∫_(-π/2)^(π/2) cos t dt = 2. So A = 2B.
3. Find B. B = ∫_(-π/2)^(π/2) [(1+A) t sin t + B t cos t] dt. Here t cos t is odd and gives 0, while t sin t is even: 2∫₀^(π/2) t sin t dt = 2[-t cos t + sin t]₀^(π/2) = 2. So B = 2(1+A).
4. Solve together. A = 2B = 4(1+A) gives A = -4/3, so 1 + A = -1/3 and B = -2/3. Hence f(θ) = -1/3 sin θ - 2/3 cos θ.

   ∫₀^(π/2) f(θ) dθ = -1/3(1) - 2/3(1) = -1

5. So. The modulus of -1 is 1.

The common slip: -1. Leaving out the modulus at the end gives -1. The question asks for the absolute value, so the answer is 1.

The idea to keep. When f appears inside a definite integral with fixed limits, that integral is just a number. Name each one, rewrite f, then substitute back to get equations for the constants.

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