lim x → π/2((∫_(x³)^((π / 2)³)(sin (2 t^(1 / 3))+cos (t^(1 / 3))) d t)/(x-π/2)²) is equal to

Solution

Maths sir · 4 steps

Both top and bottom vanish at π/2, so differentiate the integral using its variable lower limit.

1. Spot the form. At x = π/2 the lower limit x³ equals the upper limit (π/2)³, so the integral is 0. The denominator is also 0, so this is a 0/0 form and L'Hopital's rule applies.
2. Differentiate the integral. By the Leibniz rule, a variable lower limit x³ gives -f(x³) × 3x². Here f(x³) = sin(2x) + cos x, because (x³)^(1/3) = x. The denominator differentiates to 2(x - π/2).

   L = lim(x → π/2) -3x²(sin 2x + cos x)/2(x - π/2)

3. Still zero over zero. At x = π/2, sin π + cos π/2 = 0. The -3x²/2 factor goes to -3π²/8. The remaining ratio (sin 2x + cos x)/(x - π/2) tends to the derivative 2 cos 2x - sin x at π/2, which is 2(-1) - 1 = -3.
4. So. L = -3π²/8 × (-3) = 9π²/8, which is option (b).

The idea to keep. For an integral with a variable limit, differentiate with the Leibniz rule: integrand at the limit times the limit's derivative, with a minus sign for the lower limit. Then check for 0/0 again.

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