Let f(t) = ∫₀ᵗ e^(x³)( x⁸/(x⁶ + 2x³ + 2)² )dx . If f(1) + f'(1) = α e - 1/6, then the value of 150α is equal to ___________.

Solution

Maths sir · 4 steps

Put u = x³, then spot the eᵘ(g + g') pattern; the integral falls out in one line.

1. Substitute. Let u = x³, so du = 3x² dx. Write x⁸ dx = x⁶ · x² dx = 1/3u² du, and the denominator becomes (u² + 2u + 2)². The integrand is now 1/3eᵘ u²/(u²+2u+2)².
2. Spot the pattern. Take g(u) = 1/(u²+2u+2). Then g'(u) = -(2u+2)/(u²+2u+2)², so g + g' = (u²+2u+2-2u-2)/(u²+2u+2)² = u²/(u²+2u+2)². Since ∫ eᵘ(g + g')du = eᵘ g, the antiderivative is eᵘ/(u²+2u+2).

   f(t) = 1/3[e^(t³)/(t⁶+2t³+2) - 1/2]

3. Evaluate at 1. f(1) = 1/3(e/5 - 1/2) = e/15 - 1/6. By the fundamental theorem, f'(1) is the integrand at x = 1: f'(1) = (e · 1)/(1+2+2)² = e/25.
4. So. f(1) + f'(1) = e(1/15 + 1/25) - 1/6 = 8e/75 - 1/6. So α = 8/75 and 150α = 16.

The common slip: 36. Dropping the 1/3 from du = 3x² dx gives f(1) = e/5 - 1/2, so α = 6/25 and 150α = 36. The constant -1/2 not matching -1/6 is the warning sign.

The idea to keep. When eˢᵒᵐᵉᵗʰⁱⁿᵍ multiplies a messy fraction, substitute so the exponent is u, then look for g + g'. The integral of eᵘ(g + g') is just eᵘ g.

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