Let f:(0, ∞) → R and F(x)=∫₀ˣ t f(t) dt. If F(x²)=x⁴+x⁵, then Σᵣ₌₁¹² f(r²) is equal to ____________.

Solution

Maths sir · 4 steps

Differentiate both sides with the chain rule, and the integral sign gives you f directly.

1. Fundamental theorem. Since F(x)=∫₀ˣ t f(t) dt, its derivative is F'(x) = x f(x). So F'(x²) = x² f(x²).
2. Differentiate both sides. Differentiate F(x²)=x⁴+x⁵ with respect to x. The chain rule puts a factor 2x on the left side.

   x² f(x²)· 2x = 4x³ + 5x⁴ ⇒ f(x²) = 2 + 5/2x

3. Put in r. With x = r, f(r²) = 2 + 5r/2. Then the sum is Σᵣ₌₁¹²(2 + 5r/2) = 2 × 12 + 5/2 × 78 = 24 + 195.
4. So. Σᵣ₌₁¹² f(r²) = 219.

The common slip: 3562. Dropping the chain-rule factor 2x gives f(r²) = 4r + 5r², and the sum becomes 3562 instead of 219.

The idea to keep. When a function is defined by an integral with a variable limit, differentiate to remove the integral. Always multiply by the derivative of the inner limit.

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