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

Solution

Maths sir · 4 steps

This is a 0/0 form with an integral on top, so apply L'Hopital and differentiate the integral by Leibniz.

1. Check 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, which gives a 0/0 form.
2. Differentiate the integral. The variable x sits in the lower limit, so Leibniz's rule gives a minus sign and the chain factor 3x². Since (x³)^(1/3) = x, the derivative of the numerator is -cos x · 3x². The derivative of the denominator is 2(x - π/2).

   lim(x → π/2) (-3x² cos x)/2(x - π/2)

3. Apply L'Hopital again. This is still 0/0 because cos π/2 = 0. The top differentiates to -6x cos x + 3x² sin x, which at x = π/2 equals 0 + 3 × π²/4 = 3π²/4. The bottom differentiates to 2.
4. So. The limit is 3π²/4 ÷ 2 = 3π²/8, which is option (b).

The idea to keep. An integral with x in a limit becomes an ordinary function after one derivative: d/dx∫_(g(x))ᶜ f(t) dt = -f(g(x)) g'(x). Always keep the chain factor and the sign.

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