If [t] denotes the greatest integer ≤ t, then the value of 3(e - 1)/e∫₁² x²e^([x] + [x³])dx is :

Solution

Maths sir · 4 steps

On [1,2) the [x] is just 1, and x² dx is begging you to put t = x³.

1. Settle the easy part. For 1 ≤ x < 2, [x] = 1, so e^([x]) = e comes out as a constant. The single point x = 2 does not change the integral.
2. Substitute. Put t = x³, so dt = 3x² dx and t runs from 1 to 8. On each interval [k, k+1), [t] = k, which splits the integral into 7 pieces of width 1.

   ∫₁² x² e^([x]+[x³]) dx = e/3∫₁⁸ e^([t]) dt = e/3(e + e² + … + e⁷)

3. Sum the series. The geometric series is e + e² + … + e⁷ = e(e⁷ - 1)/(e - 1). So the integral equals e²(e⁷ - 1)/3(e - 1).
4. So. Multiplying by 3(e-1)/e leaves e(e⁷ - 1) = e⁸ - e, which is option (a).

The idea to keep. When a greatest integer function sits on x³ and you also have x² dx next to it, substitute t = x³ first. Then [t] is constant on each unit interval and the integral becomes a sum.

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